Class 12 Engineering Chemistry - Solutions

The mass percentage of nitrogen in histamine is _______.

Molecular formula of histamine:-

$C_5H_9N_3$

Molecular Mass

$= 111$

$\%$ Mass of Nitrogen

$=\dfrac{42}{111} \times 1000$

$= 37.84\%$

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An ideal aqueous solution containing liquid $A(M.Wt=128)$ $64$ % by weight has a vapour pressure of $145mm$ $Hg$ . If the vapour pressure of $A$ is $x$ $mm$ of $Hg$ and that of water is $155mm$ $Hg$ at the same temperature. Then find $\dfrac{x}{5}$ . The solutions is ideal.

For two miscible liquids,

$P_{total}$ = mole fraction of

$A$

$\times P_{A}^{\circ}$ + mole fraction of

$B$

$\times P_{B}^{\circ}$

Number of moles of

$A=$

$\cfrac{64}{128}=0.5$

Number of moles of

$B=$

$\cfrac{36}{18}=2$

$\Longrightarrow x=43\quad { \Lambda }_{ A }=\cfrac { 0.5 }{ 2.5 } =0.2;{ \Lambda }_{ B }=\cfrac { 2 }{ 2.5 } =0.8\\ \Longrightarrow 145=(0.2)(x)+(0.8)(155)$

$\Longrightarrow \left( \cfrac { x }{ 5 } \right) =145-\cfrac { 4 }{ 5 } \times 155=21$ mm of Hg

Does a new substance form when a solute dissolves in a solvent?

No new substance is formed when a solute dissolves in a solvent.
When a solute dissolves in a solvent a solution is formed.
Solvent dissociates the components of solute.
Example:- When $NaCl$ is dissolved in water, $H_2O$ molecules surround the negative chloride ions and positive sodium ions and split $NaCl$ molecule.

Calculate the percentage of magnesium in magnesium carbonate.

$\textbf{Step 1: Calculate the molar mass of magnesium carbonate.}$

Molar mass of magnesium carbonate ( $MgCO_3$ ) = Mass of Mg atom + Mass of C atom + Mass of 3 O atoms.

Molar mass of magnesium carbonate ( $MgCO_3$ ) = 24 + 12 + (3 $\times$ 16)

Molar mass of magnesium carbonate ( $MgCO_3$ ) = 24 + 12 + 48 = 84 g/mol

$\textbf{Step 2: calculate the percentage of magnesium in magnesium carbonate.}$

$Pecentage\ of\ magnesium = \dfrac{Mass\ of\ magnesium}{Molar\ mass\ of\ magnesium\ carbonate} \times 100$

$Pecentage\ of\ magnesium = \dfrac{24}{84} \times 100 = 28.57$ %

$\textbf{Conclusion:}$

Therefore, the percentage of magnesium in magnesium carbonate is 28.57%.

Mention names of four methods of expressing concentration of a solution.

The methods of expressing the concentration of a solution are

1. Molarity,

2. Normality,

3. Formality,

4. Percentage proportion.

Calculate percentage of oxygen present in $CO_{2}$ . [Atomic mass of O=16u. C=12u]

$\text{Step 1: Calculate the molar mass of carbon dioxide}$

$CO_2$

Atomic mass of carbon

$=12$

Atomic mass of oxygen

$=16$

Molar mass of

$CO_2=1\times 12 + 2\times 16$

$=12 + 32$

$=44 g/mol$

$\text{Step 2: Calculation of the percentage of oxygen in}$

$CO_2$

Since,

$Mass\ percentage\ of\ element=\dfrac{Mass\ of\ the\ element\ in\ the\ compound}{Molar\ mass\ of\ the\ compound}\times 100$

Therefore,

$Mass\ percentage\ of\ oxygen=\dfrac{32}{44}\times 100$

$Mass\ percentage\ of\ oxygen=\dfrac{3200}{44}$

$Mass\ percentage\ of\ oxygen=72.72\%$

Hence, the percentage of oxygen in

$CO_2$ is

$72.72\%$

$\text{Additional information:}$

$\text{ Calculation of the percentage of carbon in}$

$CO_2$

Since,

$Mass\ percentage\ of\ element=\dfrac{Mass\ of\ the\ element\ in\ the\ compound}{Molar\ mass\ of\ the\ compound}\times 100$

Therefore,

$Mass\ percentage\ of\ carbon=\dfrac{12}{44}\times 100$

$Mass\ percentage\ of\ carbon=\dfrac{1200}{44}$

$Mass\ percentage\ of\ carbon=27.27\%$

Hence, the percentage of carbon in

$CO_2$ is

$27.27\%$

At constant temperature and pressure, one mole of a pure substance free energy is identical with:

Since at constant temperature and pressure, for one mole of a pure substance, the ratio of the free energy to the chemical potential is

$1$ . Hence for one mole of a pure substance free energy is identical with chemical potential.

Van't Hoff factor for an electrolyte is greater than _____.

The van't Hoff factor for an electrolyte is greater than 1.

For example, consider

$NaCl$ which is a strong electrolyte and completely dissociates.

$NaCl \rightarrow Na^+ + Cl^-$

Thus, 1 mole of NaCl on dissociation gives 2 moles of ions.

Hence, the van't Hoff factor

$i =\dfrac {\text {Actual number of particles}}{\text {Number of particles for no ionization}}= \dfrac {2}{1}=2$ .

For strong electrolytes, van't Hoff factor 'i' equals to total number of _____ (ions/atoms) in the formula unit.

Consider the strong electrolyte

$\displaystyle A_xB_y$ .

It completely dissociates into aqueous solution.

$\displaystyle A_xB_y \rightarrow xA^{+y}+yB^{-x}$

The total number of ions in the formula unit

$\displaystyle = x+y$

The van't Hoff factor

$\displaystyle, i = \frac {\text {Actual number of particles}}{\text {Number of particles for no ionization}} = \frac {x+y}{1}=x+y$

Hence, for strong electrolytes, van't Hoff factor 'i' equals to total number of ions in the formula unit.

Van't Hoff factor for $TH(NO_3)_4$ in an aqueous solution is :

1 molecule of

$Th(NO_3)_4$ on dissociation gives fiv ions.

$Th(NO_3)_4 \rightleftharpoons Th^{4+}+4NO_3^-$
Hence, the van't Hoff factor for

$Th(NO_3)_4$ in an aqueous solution is 5.

At $25^oC$ benzene and toluene have densities 0.879 and 0.867 g/mL respectively. Assuming that benzene, toluene solutions are ideal, then the equation for the density (e) of solution as: $e=\frac {1}{100}[0.879V+0.867(100-V)]$ where V is the volume of benzene.

If true enter 1 else enter 0.

Benzene and toluene form an ideal solution.

So,

$V_{sol}=V_{ben} +V_{tol}$

mass of toluene

$= 0.867\times V_{tol}$

mass of benzene

$= 0.879\times V_{ben}$

Total mass of solution =

$0.867\times V_{tol}$ +

$0.879\times V_{ben}$

Hence, density of solution

$= \cfrac {Total \: mass}{Total \: Volume}$

$= \cfrac {0.867\times V_{tol} + 0.879\times V_{ben}}{V_{ben} +V_{tol}}$

Hence, the above given statement is incorrect.

Match column I with column II

$i$ represents the van't Hoff factor,

$\alpha$ represents the degree of dissociation and n represents the number of ions produced by complete dissociation of 1 mole of the substance or the number of simple molecules present in an associated molecule.
If association is complete

$i=\dfrac {1}{n}, \alpha=1$
If dissociation is complete

$i=x+y, \alpha=1$
If no association or dissociation

$\alpha=0, i=1$
Dependence of number of particles present after dissolution

$i > 1, i < 1$

Match column I with column II

The vapor pressure of solution is proportional to the mole fraction of solution.
The lowering in vapor pressure of solution is proportional to the mole fraction of solute.
Acetone - chloroform solution is a non ideal solution with negative enthalpy of mixing. It shows negative deviation from Raoult's law.
Hexane-ethanol solution is a non ideal solution with positive enthalpy of mixing. It shows positive deviation from Raoult's law.
The van't Hoff factor for benzoic acid in benzene is less than 1. This indicates association.
The van't Hoff's factor for glucose in water is equal to unity. This indicates neither association nor dissociation.
The van't Hoff factor for

$NaCl$ in water is greater than unity. This indicates dissociation.

For weak electrolytes, van't Hoff factor 'i' is a measure of the degree of dissociation of weak electrolyte.
If true enter 1, if false enter 0.

For weak electrolytes, van't Hoff factor 'i' is a measure of the degree of dissociation of weak electrolyte.
Higher is the value of the van't Hoff factor, higher will be the degree of dissociation.
The relationship between the van't Hoff factor i and the degree of dissociation

$\alpha$ is

$\alpha = \frac {i-1}{n-1}$
Here n is the number of ions produced by the complete dissociation of 1 mole of substance.

Van't Hoff factor of a mixture of two moles of KI with 1 mole $HgI_2$ in a solution of water is :

Two moles of KI react with on mole of

$HgI_2$ to form one mole of the complex

$K_2[HgI_4]$ .

$2KI + HgI_2 \rightleftharpoons K_2[HgI_4]$

In aqueous solution, one mole of the complex

$K_2[HgI_4]$ dissociates to give 3 moles of ions.

$K_2HgI_4 \rightleftharpoons 2K^+ + [HgI_4]^{2-}$

Van't Hoff factor

$i=\dfrac {n(observed)}{n(theoretical)} = \dfrac {3}{1}=3$

Thus, Van't Hoff factor of a mixture of two moles of KI with 1 mole

$HgI_2$ in a solution of water is 3.

What is the difference between lyophobic sol and lyophilic sol ?

The difference between lyophobic and lyophilic sols is as given below.

Lyophobic sols	Lyophilic sols
Disperse phase and dispersion medium have low affinity for each other.	Disperse phase and dispersion medium have strong affinity for each other.
The dispersion medium and the sol have same viscosity.	Sol has higher viscosity than dispersion medium.
The dispersion medium and the sol have same surface tension.	Sol has lower surface tension than dispersion medium.
These sols are irreversible and less stable.	These sols are reversible and highly stable.

How is the vapour pressure of a solvent affected when a non-volatile solute is dissolved in it ?

When a non volatile solute is added to a solvent, the vapour pressure of the resulting solution is lower than the vapour pressure of pure solvent.
This is because, the surface area is partly occupied by non volatile solute molecules and partly by solvent molecules. Due to this, the rate of evaporation is lowered.

For an ideal solution, ${\Delta}_{mix}H$ is _______ .

Ideal Solutions : A solution which obeys Rault's law exactly at all concentrations and at all temperatures is called an ideal solution.
An ideal solution possesses the following characteristics :
(i)Volume change of mixing shoule be zero.
V solvent + V solute = V solution
(ii)Heat change on mixing should be zero.

For an ideal solution, ${\Delta }_{mix}V$ is

The vapour pressure of water is $12.3\ kPa$ at $300\ K$ . Calculate the vapour pressure of $1$ molal solution of a non-volatile solute in it.

The vapour pressure of water is

$12.3$ kPa at

$300$ K. We need to calculate the vapour pressure of

$1$ molal solution of a non-volatile solute in it.

$1000$ g of water contains

$1$ mole of solute.

The molar mass of water is

$18$ g/mol.

Number of moles of water

$\displaystyle = \frac {1000}{18} = 55.56$ mol

Mole fraction of the solute in the solution is

$\displaystyle x = \frac {1}{1+55.56} = 0.0177$ .

The relative lowering in the vapour pressure is equal to the mole fraction of solute.

$\displaystyle \dfrac {p^0-p}{p^0} =x$

$\dfrac {12.3-p}{12.3} = 0.0177$

$\Rightarrow p = 12.08$

Hence, the vapour pressure of the solution is

$12.08$ kPa.

Vapour pressure of water at $293\ K$ is $17.535\ mm\ Hg$ . Calculate the vapour pressure of water at $293\ K$ when $25\ g$ of glucose is dissolved in $450\ g$ of water.

The molar masses of glucose and water are

$180$ g/mol and

$18$ g/mol respectively.

Number of moles of glucose

$\displaystyle = \frac {25}{180} = 0.139$

Number of moles of water

$\displaystyle = \frac {450}{18} = 25$

Mole fraction of glucose

$\displaystyle = \frac {0.139}{0.139+25} = 0.0055$

Vapour pressure lowering is directly proportional to the mole fraction

$\displaystyle \frac {P^0-P}{P^0} = X$

$\dfrac {17.535-P}{17.535} = 0.0055$

$P = 17.438$

Hence, the vapour pressure of a solution is

$17.438$ mm Hg.

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain.
Cyclohexane, $KCl,\ CH_{3}OH,\ CH_{3}CN$

n-Octane is non polar and can dissolve non-polar solutes. It cannot dissolve polar (and ionic) solutes.

$\text{Cyclohexane}$ is non-polar. Hence, easily soluble in n-octane.

$\text{Methanol}$ and

$\text{acetonitrile}$ are polar and have very low solubility in n-octane.

$\text{KCl}$ is ionic compound and hence, insoluble in n-octane.

The increasing order for solubility in n-octane is as follows:

$\displaystyle KCl < CH_3OH < CH_3CN < \text{Cyclohexane}$

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

$\bf{Explanation:}$

A solution is formed when two or more components are mixed together.

The component which is present in a large amount in solution than others is called a solvent and the component which is less in amount in a solution is called the solute.

There are

$9$ types of solution, which are as follow:

$Solution\ type$	$Solute$	$Solvent$	$Example$
Gas in Gas	Gas	Gas	Air
Gas in Liquid	Gas	Liquid	Soda water
Gas in Solid	Gas	Solid	$H_2$ in $Pt$
Liquid in gas	Liquid	Gas	Water vapor in the air
Liquid in Liquid	Liquid	Liquid	Alcoholic beverages.
Liquid in Solid	Liquid	Solid	Water into ice
Solid in Gas	Solid	Gas	Camphor in air
Solid in Liquid	Solid	Liquid	Salt in water
Solid in Solid	Solid	Solid	Metal alloys, Bronze

Define an ideal solution and write one of its characteristics.

An ideal solution is the solution which obeys Raoult's law exactly over entire range of concentration. Such solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecualr forces.
The characteristics of ideal solutions:
(1) It must obey Raoult's law.
(2) The enthalpy of mixing should be zero.
(3) The volume of mixing should be zero.

Arrange the following compounds in increasing order of solubility in water:
${ C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2 },{ \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }NH,{ C }_{ 2 }{ H }_{ 5 }N{ H }_{ 2 }$

The increasing order of solubility in water is

$\displaystyle { C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2 } <{ \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }NH <{ C }_{ 2 }{ H }_{ 5 }N{ H }_{ 2 }$ .
The lower aliphatic amines are soluble in water because they are capable of forming hydrogen bonds with water. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part.
The extent of hydrogen bonding in secondary alcohols is lower than the extent of hydrogen bonding in primary amines.
Hence, the solubility of secondary amines is lower than the solubility of primary amines.
Aromatic amines are water insoluble due to larger hydrocarbon part which tends to retard the formation of hydrogen bonds.

A 1.00 molal aqueous solution of trichloroacetic acid $(CCl_3COOH)$ is heated to its boiling point. The solution has the boiling point of $100.8^oC$ . Determine the van 't Hoff factor for trichloroacetic acid. $(K_b$ for water 0.512 K kg $Mol^{-1}$ )
OR
Define the following terms:
(i) Mole fraction
(ii) Isotonic solutions
(iii) Van 't Hoff factor
(iv) Ideal solution

Molarity of solution

$= m= 1.00 m$

Boiling point of solution

$= T_b= 100.8 ^oC = 373.8 \ K$

Boiling point of water

$= T_b^o = 100.0^o C= 373 \ K$

Observed elevation in boiling point

$= 373.8- 373 = 0.8 \ K$

$K_b$ for water

$= 0.512 K kg Mol^−$

Calculated elevation in boiling point

$\Delta T_b= K_b \times m = 0.512 \times 1 = 0.512 \ K$

Van't Hoff factor

$i = \dfrac {Observed \ colligative \ property}{Calculated \ colligative \ property}$

$\Rightarrow i= \dfrac {0.8}{0.512}= 1.56$

(i) Mole fraction= The ratio of the number of moles of a given component of a mixture to the total number of moles of all the components.

$Mole \ fraction = \dfrac {Number \ of \ moles \ of \ the \ component}{Total \ Number \ of \ moles \ of \ all \ the \ components}$

(II) Isotonic solution= An isotonic solution refers to two solutions having the same osmotic pressure across a semipermeable membrane. This state allows for the free movement of water across the membrane without changing the concentration of solutes on either side.

(iii) The van 't Hoff factor is a measure of the effect of a solute upon colligative properties such as osmotic pressure, relative lowering in vapor pressure, elevation of boiling point and freezing point depression.

Van't Hoff factor

$i = \dfrac {Observed \ colligative \ property}{Calculated \ colligative \ property}$

(iv) Ideal Solution= Ideal solution, homogeneous mixture of substances that has physical properties linearly related to the properties of the pure components. An ideal solution is a solution with thermodynamic properties analogous to those of a mixture of ideal gases. The enthalpy of mixing is zero. The vapor pressure of the solution obeys Raoult's law, and the activity coefficient of each component is equal to one.

Vapour pressure of three liquids A, B, C are in their ascending order. Identify the liquid that has least critical temperature in its gaseous state.

The temperature at and above which a substance cannot be liqueified no matter however high the applied pressure may be is known as the critical temperature. A liquid having the least vapor pressure will have the highest boiling point and hence the least critical temperature. Hence the vapor pressure is directly proportional to the critical temperature and this gives A having the least critical temperature in the gaseous state

A mixture containing four miscible liquids P, Q, R, and S are subjected to fractional distillation. Liquid P is distilled off first, into the receiver flask. Then S, followed by R and Q is left behind in the distillation flask. Based on the above observations, arrange the liquids in ascending order of their vapour pressures.

Distillation is a process that is used to separate liquids having different boiling point. Higher the vapor pressure, lower is the intermolecular forces and hence lower will be the boiling point. Hence the substance having the highest vapor pressure will be distilled out first. Since the order of liquids being distilled out is

$Q<R<S<P$ , this would also be the ascending order of vapor pressure of the liquids

Why is be solubility of $NaCl$ in water $311\,g\, l^{-1}$ $N$ but is zero in benzene?

$NaCl$ is an ionic compound and soluble only in the polar solvent because we know that "like dissolve like", water is a polar solvent and benzene is a non-polar solvent,

hence, the solubility of NaCl in water is

$311 gl^{-1}$ and zero in benzene.

What is the value of $\Delta H_{mix}$ for an ideal solution ?

For an ideal solution, their is no change in the bonding interactions before and after mixing of solution;

so, the enthalpy value remains same

and we get

$\Delta H_{mix}$ = 0

a. Which gas is dissolved in soft drinks?
b. What will you do to increase the solubility of this gas ?

a. Carbon dioxide gas is dissolved in soft drinks. It adds acidic flavour and pleasantly mouth feel.

b. As you can see in the figure the solubility of gas increases with the increases in the pressure. It follows the Henry's law. The law is stated as: " The solubility of a gas in a liquid is directly proportional to the pressure of the gas at the surface of the solution"

What happens to vapour pressure of water if a tablespoon of sugar is added to it?

The vapour pressure of water is lowered if a tablespoon of sugar is added to it. When a non volatile solute is added to a solvent, the vapour pressure of the resulting solution is less than the vapour pressure of pure solvent.

What will be the value of van't Hoff factor $\left(i\right)$ of benzoic acid if it demerises in aqueous solution? How will the experimental molecular weight vary as compared to the normal molecular weight?

$i=\dfrac { Particles\quad after\quad association }{ Particles\quad before\quad association }$

$=\dfrac { 1 }{ 2 } =0.5$
van't Hoff factor

$\left( i \right) =\dfrac { Normal\quad molecular\quad weight }{ Experimental\quad molecular\quad weight }$

What will be the value of Van't Hoff factor for ethanoic acid in benzene?

In benzene, two molecules of ethanoic acid associate to form a dimer.

$\displaystyle 2CH_3COOH \rightleftharpoons (CH_3COOH)_2$
The van't Hoff factor

$\displaystyle i = \dfrac {\text {number of solute particles in solution}}{\text {theoretical number of solute particles for no association}}$

$\displaystyle i=\dfrac {1}{2}=0.5$

Derive van't Hoff general solution equation.

Consider a dilute solution of a volume

$V$ (in

$dm^3$ ) containing W grams of a substance having molecular weight

$M$ at an absolute temperature

$T$ . Then, concentration of the solution, C

$=$ n

$/$ V and No. of moles of the substance, n

$=$ W

$/$ M.
By van't Haff. Boyle's law, at a constant temperature, the osmotic pressure of a solution is directly proportional to its molar concentration or inversely proportional to the volume of the solution. Thus,

$\pi \alpha$ C(when temperature is constant)
Van't Hoff Charle's law at the constant concentration the osmotic pressure of a dilute solution is directly proportional to the absolute temperature(

$T$ ) i.e.

$\pi\alpha$ T

$\pi\alpha$ CT

$v=CRT$ .
The equation is called van't Hoff general solution equation.

$R$ is constant of proportionality, also called general solution constant or gas constant.

What is relative lowering of vapour pressure? How is it useful to determine the molar-mass of a solute?

As the solute molecules are non-volatile, the vapour above the solution consists of only solvent molecules. After adding the solute, the vapour pressure of the solution is found to be lower than that of the pure liquid at a given temperature. This is known as relative lowering of vapour pressure.

The relative lowering of vapour pressure

$\displaystyle = \dfrac {\Delta p}{p^0_1}= \dfrac { p_1^0-p}{p^0_1}$
Here,

$\displaystyle p^0_1$ is the vapour pressure of pure solvent and

$\displaystyle \Delta p$ is the vapour pressure lowering of solution.

To determine the molar-mass of a solute, the following expression is used

$\displaystyle \dfrac {\Delta p}{p^0_1} = \dfrac {W_2M_1}{W_1M_2}$

$\displaystyle W_1$ and

$\displaystyle W_2$ are the masses of solvent and solute respectively.

$\displaystyle M_1$ and

$\displaystyle M_2$ are the molar masses of solvent and solute respectively.

The vapour pressure of pure benzene at a certain temperature is $0.850$ bar. A non-volatile, non-electrolyte solid weighing $0.5$ $g$ when added to $39$ $g$ of benzene (molar mass $78$ $g$ ${mol}^{-1}$ ), vapour pressure becomes $0.845$ bar. What is the molar mass of the solid substance?

$\displaystyle \dfrac {p^0-p}{p^0}=\dfrac {W_2M_1}{W_1M_2}$

$\displaystyle p^0 = 0.850 \: bar =$ vapour pressure of pure benzene.

$\displaystyle p = 0.845 \: bar =$ vapour pressure of solution.

$\displaystyle W_1 = 39 \: g=$ mass of benzene

$\displaystyle W_2 = 78 \: g=$ mass of solute

$\displaystyle M_1 = 78 \: g/mol =$ Molar mass of benzene.

$\displaystyle M_2 =$ Molar mass of solute.

$\displaystyle \dfrac {0.850 \: bar-0.845 \: bar}{0.850 \: bar}=\dfrac {0.5 \: g \times 78 \: g/mol}{39 \: g \times M_2}$

$\displaystyle M_2 = 170 \: g/mol$

'The extent to which a solute is dissociated or associated can be expressed by Van 't Hoff factor'. Substantiate the statement.

The ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass,known as Van't Hoft factor.
So,we can say that the extent to which a solute is dissociated or associated can be expressed by Van't Hoft factor.
This is represented by $i$ :
$i=\dfrac { observed\quad no.\quad of\quad solute\quad particles }{ Theoretical\quad no.\quad of\quad solute\quad particles }$
If $i >1$ ,the molecule are said to be dissociated.
If $i=1$ ,the molecule have neither associated nor dissociated.
If $i <1$ ,the molecule are said to be associated.

Example:

$A_xB_y \rightleftharpoons xA^{y+} \quad yB^{x-}$

$t=0 \quad C \quad 0 \quad 0$

$t=t_f \quad C(1-\alpha) \quad xC\alpha \quad yC\alpha$ (

$\alpha$ is the degree of dissociation)

SO ,

$i=\dfrac{C(1-\alpha)+xC\alpha+yC\alpha}{C}$

$\Rightarrow (1-\alpha)+x\alpha+y\alpha$

$i=1+(x+y-1)\alpha$ (Here (x+y) are the total particles produced by one molecule of solute)

Derive the relationship between relative lowering of vapour pressure and molar mass of solute.

The expression for the relative lowering of vapour pressure is

$\displaystyle \dfrac {\Delta P}{P^0} = \dfrac {P^0-P}{P^0} =X_2$ .....(1)
Here,

$\displaystyle \dfrac {\Delta P}{P^0}$ and

$\displaystyle \dfrac {P^0-P}{P^0}$ represents relative lowering of vapour pressure and

$\displaystyle X_2$ represents mole fraction of solute.

But the mole fraction of solute

$\displaystyle X_2 = \dfrac {n_2}{n_1+n_2} = \dfrac {\dfrac {W_2}{M_2}}{\dfrac {W_2}{M_2}+\dfrac {W_1}{M_1}}$ .....(2)
Here,

$\displaystyle W_1$ and

$\displaystyle M_1$ are the mass and molar masses of solvent and

$\displaystyle W_2$ and

$\displaystyle M_2$ are the mass and molar masses of solute.

$\displaystyle n_1$ and

$\displaystyle n_2$ are the number of moles of solute and solvent respectively.

For dilute solutions

$\displaystyle n_1>>n_2$
So

$\displaystyle n_2$ may be neglected in comparison with

$\displaystyle n_1$ .
Hence, equation (2) becomes

$\displaystyle X_2 = \dfrac {n_2}{n_1} = \dfrac {\dfrac {W_2}{M_2}}{\dfrac {W_1}{M_1}}$ .....(3)

From equation (1) and equation (3)

$\displaystyle \dfrac {\Delta P}{P^0} = \dfrac {P^0-P}{P^0} = \dfrac {\dfrac {W_2}{M_2}}{\dfrac {W_1}{M_1}}$
This gives the relationship between relative lowering of vapour pressure and molar mass of solute.

Alkaline solution of $Hg{Cl}_{2}$ and $KI$ is called _________.

Potassium tetraiodomercurate(II) is an inorganic compound consisting of potassium cations and the tetraiodomercurate(II) anion. It is mainly used as Nessler's reagent, a 0.09 mol/L solution of potassium tetraiodomercurate(II) (

${ K }_{ 2 }[Hg{ I }_{ 4 }]$ ) in 2.5 mol/L potassium hydroxide, used to detect ammonia.

Mention the enthalpy of mixing $({\triangle}_{mix}H)$ value to form an ideal solution.

$\Delta _{mixing}H$ value to form the ideal solution is zero.

Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquids are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure, in bar, of a mixture of 25 g heptane and 35 g of octane?

(A) Heptane

$C_7H_{16} m_A=100$
(B) Octane

$C_8H_{18} m_B=114$

$n_A=\dfrac{w_A}{m_A}=\dfrac{25}{100}=0.25; n_B=\dfrac{35}{114}=0.3$

$x_A=\dfrac{0.25}{0.25+0.30}=0.45; x_B=\dfrac{0.3}{0.25+0.30}=0.55$

$p=p^0_Ax_A+p^0_Bx_B$

$=105.2\times 0.45+46.8\times 0.55$

$=47.34+25.74 = 73.08 kPa$

An aqueous solution containing $28\%$ by mass of a liquid A (mol. mass = $140$ ) has a vapour pressure of $160$ mm of $37^o C$ . Find the vapour pressure of the pure liquid A (The vapour pressure of water at $37^oC$ is $150$ mm).

For two miscible liquids,

$P_{total}=$ Mole fraction of

$A\times p_A^0 +$ Mole fraction of

$B\times p^0_B$

No. of moles of

$A=\dfrac{28}{140}=0.2$

Liquid B is water. Its mass is (100-28), i.e., 72.

No. of moles of

$B=\dfrac{72}{18}=4.0$

Total no. of moles

$= 0.2 + 4.0 = 4.2$

Given,

$P_{total}=160 mm$

$p_B^0 = 150 mm$

So,

$160=\dfrac{0.2}{4.2}\times p^0_A+\dfrac{4.0}{4.2}\times 150$

$p^0_A=\dfrac{17.14\times 4.2}{0.2}=360\ mm$

Potassium ferrocyanide is 50% ionised in aqueous solution. Its van't Hoff factor will be:

1142847_666120_ans_3dccbaaf47b747cb96c69cd9228aad0e.jpg

A solution containing $8.44 g$ of sucrose in $100 g$ of water has a vapour pressure $4.56 mm$ of $Hg$ at $273 K$ . If the vapour pressure of pure water is $4.58 mm$ of $Hg$ at the same temperature, calculate the molecular weight of sucrose.

Let M g/mol be the molecular weight of sucrose. The number of moles of sucrose

$\displaystyle = \dfrac {8.44}{M}$ .

The molecular weight of water is 18 g/mol. The number of moles of water

$\displaystyle = \dfrac {100}{18}=5.56$ .

The mole fraction of sucrose

$\displaystyle = \dfrac { \dfrac {8.44}{M}}{\dfrac {8.44}{M} + 5.56 } \simeq \dfrac {8.44}{M \times 5.56} \simeq \dfrac {1.52}{M}$

The relative lowering of vapour pressure is equal to the mole fraction of solute.

$\displaystyle \dfrac {P^0-P}{P^0} = \chi_{solute}$

$\displaystyle \dfrac {4.58-4.56}{4.58}=\dfrac {1.52}{M}$

$\displaystyle M=348$ g/mol.

Hence, the molecular weight of sucrose is

$348 g/mol$ .

Calculate the vapour pressure of an aqueous solution of 1.0 molal glucose solution at $100^oC.$

Molality

$=\dfrac{w}{m\times W}\times 1000$
where, w = mass of solute in grams;
W = mass of solvent in grams

$1.0=\dfrac{w}{m\times W}\times 1000$
or

$\dfrac{w}{m\times W}=\dfrac{1.0}{1000}=0.001$
Applying Raoult's law for dilute solution,

$\dfrac{P_0-P_s}{P_0}=\dfrac{w}{m\times W}\times M (M=18)$

$\dfrac{760-P_s}{760}=0.001\times 18 (P_0=760 mm at 100^oC)$
or

$P_s=760-760\times 0.001\times 18$

$=760-13.68$

$=746.32 mm$

The vapour pressure of two pure liquids A and B forming an ideal solution are 400 mm Hg and 800 mm Hg respectively at temperature I. A liquid mixture containing 3:1 molar composition of A and B is present in a cylinder closed by a piston so that the pressure can be varied. The solution is slowly vaporised at temperature I by decreasing the applied pressure from a starting of 760 mm Hg. A pressure gauge (in mm Hg) is connected

Vapour pressure of pure liquid

$(P^oA)=400\ mmHg$

mole fraction of

$A=3/4$

Partial vapour pressure of

$A(PA)=P^oA\times x_A$

$\Rightarrow \dfrac{3}{4}\times 400$

$\Rightarrow \boxed{300\ mmHg}$

vapour pressure of pure liquid

$B(P^oB)=800mmHg$

mole fraction of

$B=1/4$

Partial vapour pressure of

$B(P_B)=P_B^o\times x_B$

$=800\times \dfrac{1}{4}$

$=\boxed{200\ mmHg}$

Total partial pressure

$P_T=300+200$

$=500\ mmHg$

Hence, pressure gauge is connected to

$500\ mmHg$ .

A cylinder fitted with an air tight piston contains a small amount of a liquid at a fixed temperature. The piston is moved out so that the volume increases
(a) What will be the effect on the change of vapour pressure initially?
(b) How will the rates of evaporation and condensation change initially?
(c) What will happen when equilibrium is restored finally and what will be the final vapour pressure?

(a) If the vol of the container is suddenly increased, them

the vapour pressure would decrees initially Therefore

Same amount of vapour is distributed in a large volume

(b) at constant temperature, rate of evaporation is constant

and rate of condensation decrees initially

rate of evaporation

$=$ rate of condensation

and final vapour pressure

$=$ original vapour

pressure of system.

1115831_789105_ans_3b4bacc1cf0b477baa2c3f18282df98d.jpg

The freezing point of a solution containing 5.85 g of $NaCl$ in 100g of water is $-3.348^o$ C. Calculate van't Hoff factor 'i' for this solution. What will be the experimental molecular weight of NaCI?
( $K_f$ for water = $1.86 K kg mol^{-1}$ , at. wt. Na = 23, Cl = 35.5)

$w_{ NaCl }$ =5.85 g;

$w_{ H_2O }$ = 100 g;

$\Delta T_f$ = 3.348 K;

$K_f$ =1.86 Kkg/mol

As,

$\Delta { T }_{ f }=i{ K }_{ f }m\\ m=\frac { { w }_{ NaCl } }{ { M }_{ NaCl } } \times \frac { 1000 }{ { w }_{ { H }_{ 2 }O } } \\ m=\frac { 5.85 }{ 58.5 } \times \frac { 1000 }{ 100 } =1\\Therefore,\ i=\frac { 3.348 }{ 1.86 } =1.8\\ Also,\ i=\frac { { Normal\ molecular\ wt } }{ { Abnormal\ molecular\ wt } } \\ 1.8=\frac { 58.5 }{ { M } } \\ \therefore M=32.5\ g/mol$

Vant Hoff factor is

$1.8$ and experimental molecular weight of

$NaCl$ is 32.5 g/mol.

Two moles of $O_2$ gas is collected over water at 400 K temperature in 2 litre vessel. If the pressure of dry $O_2$ gas is 32.20 bar then find the vapour pressure of water under the same conditions.

For

$\displaystyle O_2$ gas collected over water, the pressure

$\displaystyle P= \dfrac {nRT}{V}$

$\displaystyle P=\dfrac { 2 \ mol \times 0.08206 \ L \cdot atm / mol \cdot K \times 400 \ K} {2 \ L}$

$\displaystyle P=32.824 \ atm$

Convert unit from atm to bar

$\displaystyle P=32.824 \ atm \times \dfrac {1.01325 \ bar}{1 \ atm}$

$\displaystyle P=33.26 \ bar$

The pressure of dry

$\displaystyle O_2$ gas is 32.20 bar.

The vapour pressure of water

$\displaystyle = 33.26 \ bar - 32.20 \ bar =1.06 \ bar$

The vapour pressure of a pure liquid at $25$ is 100 mm $Hg$ . Calculate the relative lowering of vapour pressure if the mole fraction of solvent in solution is 0.8.

$P^0=100mm$

$Hg$

Relative lowering of vapor pressure

$(P-P^0)=x\times P^0=0.8\times 100 mm\text{ } Hg=80mm\text{ } Hg$

Therefore, the relative lowering of vapor pressure is

$80$ mm of

$Hg.$

Identify the solute & the solvent in the following solution:
a) sugar solution
b) Air
c) Rain water
d) Aerated drinks

Solution	Solute	Solvent
Sugar solution	Sugar	Water
Air	Oxygen and other gas	Nitrogen
Grains	Salt,Clay,Air bubbles	Water
Aerated drinks	$CO_2$	Water

A mixture of two miscible liquids A and B is distilled under equilibrium conditions at $1$ atm pressure. The mole fraction of A in solution and vapour phase are $0.30$ and $0.60$ respectively. Assuming ideal behaviour of the solution and the vapour. Calculate the ratio of the vapour pressure of pure A to that of pure B.

Let us consider the problem:

In a Solution

${x_a} = 20\,\,\,\,{x_b} = 0.70$

In a vapour phase,

$x_A^\prime = 0.60;x_B^\prime = 0.40$

(Using Dalton's and raoult's law)

$x_A^\prime = 0.60 = \frac{{{P_A}}}{P} = \frac{{{P_A}}}{{{P_A} + {P_B}}} = \frac{{0.30p_A^o}}{{0.30p_A^o + 0.70p_B^o}}$

Implies that

$x_B^\prime = \frac{{0.70p_B^o}}{{0.30p_A^o + 0.70p_B^o}}$

Implies that

$\frac{{x_A^\prime }}{{x_B^\prime }} = \frac{{0.60}}{{0.40}} = \frac{{0.30p_A^o}}{{0.70p_B^o}}$

Hence

$\frac{{p_A^o}}{{p_B^o}} = \frac{{0.60 \times 0.70}}{{0.40 \times 0.30}} = 3.5$ .

Define an ideal solution.

The ideal solution is a Homogeneous mixture of substances that has physical properties linearly related to the properties of the pure components.
The classic statement of this condition is Raoult's law, which is valid for many highly dilute solutions and for a limited class of concentrated solutions, namely, those in which the interactions between the molecules of solute and solvent are the same as those between the molecules of each substance by itself.
Solutions of benzene and toluene, which have very similar molecular structures, are ideal: any mixture of the two has a volume equal to the sum of the volumes of the separate components, and the mixing process occurs without absorption or evolution of heat. The vapour pressure of the solutions is mathematically represented by a linear function of the molecular composition.

At $90^o$ C, the vapour pressure of toluene is $400$ Torr and that of o-xylene is $150$ Torr. What is the composition of the liquid mixture that boils at $90^o$ C when the pressure is $0.50$ atm? What is the composition of the vapour produced?

Given that

vapour pressure of toluene,

$p_{toluene}=400$ Torr

vapour pressure of o-xylene,

$p_{o-xylene}=150$ Torr at

$90^o$ C

let mixute has mole fraction of toluene as

$\chi_{t}$ and that of o-xylene is

$\chi_{o}$

pressure of the mixture

$P_T=0.5 atm=0.5 \times 760 =380\ torr$

using Raolt's law

$P_T=\chi_t.p_{toluene}+\chi_o.p_{o-xylene}$

since

$\chi_t=1-\chi_o$

$380=\chi_t.400+(1-\chi_t)150$

$\chi_t=\frac{380-150}{250}=0.92$

and

$\chi_o=1-0.92=0.08$

molefraction of toluene is 0.92 and that of o-xylene is 0.08

Calculate the following.
The vapour pressure of the solution at $298$ K.

According to Raoult's law

$P_{s}=\chi_{solvent}P_0$

where

$P_s$ = vapour pressure of solution

$\chi_{solvent}$ = mole fraction of solvent

$P_0$ =vapour pressure of solvent=0.024 atm (given)

$\chi_{solvent}=\frac{n_{solvent}}{n_{solvent}+n_{solute}}$

solvent is

$H_2O (M_{H_2O}=18g/mol)$ and solute is sucrose

$C_{12}H_{22}O_{11} (M_{C_{12}H_{22}O_{11}}=342g/mol)$

n=mass/molar mass

given that solution contains 68.4 g sucrose and 1000 g water.

Thus

${ n }_{ H_{ 2 }O }=\frac { 1000 }{ 18 } =55.55\\ { n }_{ C_{ 12 }H_{ 22 }{ O }_{ 11 } }=\frac { 68.4 }{ 342 } =0.2\\ \chi _{ { H }_{ 2 }O }=\frac { 55.55 }{ 0.2+55.55 } =0.9964\\ { P }_{ solution }= 0.9964 \times 0.024$

${ P }_{ solution }=0.0239\ atm$

At $25^o$ C the saturated vapour pressure of water is $3.165$ kPa ( $23.75$ mm Hg). Find the saturated vapour pressure of a $5\%$ aqueous solution of urea (carbamide) at the same temperature. (Molar mass of urea $=60.05$ g $mol^{-1}$ )

Let us consider the problem:

Given that:

$W = 5g,{W_A} = 95g,{M_B} = 60.05gmo{l^{ - 1}}$

${M_A} = 18gmo{l^{ - 1}},{{\rm{P}}^{\rm{0}}}{_{\rm{A}}} = 3.165kPa$

Putting the values in expression

$\dfrac{P^o_{A}- P}{P^o_{A}}= \dfrac{{{W_B} \times {M_A}}}{{{M_B} \times {W_A}}}$

$\dfrac{{3.165kPa - P}}{{3.165kPa}} = \dfrac{{5g \times 18gmo{l^{ - 1}}}}{{60.05gmo{l^{ - 1}} \times 95g}} = 0.015$

Hence,

$P = 3.165kPa - 0.015 \times 3.165kPa$

$P = 3.118kPa$ .

The vapour pressure of an aqueous solution of glucose is 750mm at 373K. find molality and mole fraction

$373K$ V.P of water is

$750 mm Hg$

We know that,

$\dfrac{(P_o-P)} { P_o} = \dfrac{w\times M} { m\times W}$

$molality = \dfrac{(P0-P)} { P0} \times \dfrac{1000}{M}$

$=\dfrac{(760-750)} { 760} \times \dfrac{ 1000}{18}$

$= 0.73 m$

Mole fraction

$= \dfrac{(P0-P)} { P0} = \dfrac{10}{760} = 0.0131$

Identify the solute & the solvent in the following solutions:
(a) Sugar solution (b) Air (c) Grains (d) Aerated drinks.

A simple solution is basically two substances that are evenly mixed together. One of them is called the solute and the other is the solvent. A solute is a substance to be dissolved. The solvent is the one doing the dissolving.

a) Solute-Sugar and Solvent-Water
b) Solute-Oxygen and all other minor components and Solvent-Nitrogen
c) Grains are solute hence solvent can't be defined for a solute.
d} There is no solute and solvent in aerated drinks because it is a colloid and for colloid, the solute is called dispersed phase and the solvent is called dispersion medium, and the dispersed phase in the aerated drink is carbon dioxide and the dispersion medium is water.

Derive the relationship between relative lowering of vapour pressure and mole fraction of the volatile liquid.

As we know that, for any solution, the partial pressure of each volatile component in the solution is directly proportional to its mole fraction.

$\therefore {p}_{A} = {x}_{A}$ ; where

${p}_{A}$ is vapour pressure of solvent having mole fraction

${x}_{A}$ .

${p}_{A} = {{p}^{0}}_{A} \times {x}_{A} ..... \left( 1 \right)$

But,

${x}_{A} + {x}_{B} = 1 \Rightarrow {x}_{A} = 1 - {x}_{B}$ , where

${x}_{B}$ is mole fraction of non-volatile solute

$B$

Now from

${eq}^{n} \left( 1 \right)$ , we have

${p}_{A} = {{p}^{0}}_{A} \left( 1 - {x}_{B} \right)$

$\Rightarrow {p}_{A} = {{p}^{0}}_{A} - {{p}^{0}}_{A} {x}_{B}$

As we know that, total vapour pressure

$\left( p \right)$ is-

$p = {p}_{A} + {p}_{B}$

As non-volatile does not have any vapour pressure.

$\therefore {p}_{B} = 0$

$\therefore p = {p}_{A} ..... \left( 2 \right)$

$\Rightarrow p = {{p}^{0}}_{A} - {{p}^{0}}_{A} {x}_{B}$

$\Rightarrow {{p}^{0}}_{A} {x}_{B} = {{p}^{0}}_{A} - p$

$\Rightarrow {x}_{B} = \cfrac{{{p}^{0}}_{A} - p}{{{p}^{0}}_{A}}$

As,

$p = {p}_{A} \; \left( \text{From } \left( 2 \right) \right)$

$\therefore {x}_{B} = \cfrac{{{p}^{0}}_{A} - {p}_{A}}{{{p}^{0}}_{A}}$

Hence the relationship between relative lowering of vapour pressure and mole fraction of the volatile liquid is-

${x}_{B} = \cfrac{{{p}^{0}}_{A} - {p}_{A}}{{{p}^{0}}_{A}}$

Out of two $0.1$ molal solutions of glucose and of potassium chloride, which one will have a higher boiling point and why?

Given that the concentration of both the solution (potassium chloride and glucose) is same than 0.1 molal solutions of potassium chloride will have a greater boiling point elevation than glucose solution as it depends on the number of ions present in solution. Since

$NaCl$ is an ionic compound it will produce 2 ions in the solution while glucose is a covalent compound and it will remain as such.

$400\ g$ of $20\%\ (w/w)$ solution is cooled. $50\ g$ of solute is precipitated. What is the %(w/w) of solute in the remaining solution?

Weight of solution = 400g

Weight of solute = $\cfrac { 20\times 400 }{ 100 }$ = 80 g

Weight of solute precipitated = 50g

Weight of solute remained in solution = 80-50 = 30g

Weight of solution remained = 400 – 50 = 350g

Hence % by weight = $\cfrac { 30 }{ 350\times 100 }$ = 8.57%

Vapour pressure of liquids $A$ and $B$ at $298\ K$ is $300\ mm\ Hg$ and $450\ mm\ Hg$ . Calculate the mole fraction of $A$ in the mixture.

[Given total vapour pressure of solution $=405\ mm\ Hg$ .]

It is given that,

${ P }^{ 0 }_{ A }=300$ mm of

$Hg$

${P }^{ 0 }_{ B }=450$ mm of

$Hg$

${P }_{ Total }=405$ mm of

$Hg$

From Raoults law, we have

${P }_{ total }={ P }_{ A }+{ P }_{ B }$

${P }_{ total }={ P }^{ 0 }_{ A }{ X }_{ A }+{ P }^{ 0 }_{ B }\left( 1-{ X }_{ A } \right)$

${ P }_{ total }={ P }^{ 0 }_{ A }{ X }_{ A }+{ P }^{ 0 }_{ B }-{ P }^{ 0 }_{ B }{ X }_{ A }$

${ P }_{ total }=\left( { P }^{ 0 }_{ A }-{ P }^{ 0 }_{ B } \right) { X }_{ A }+{ P }^{ 0 }_{ B }$

$405=\left( 300-450 \right) { X }_{ A }+450$

$-45=-150{ X }_{ A }$

${X }_{ A }=0.3$
Therefore,

${ X }_{ B }=1-{ X }_{ A }$

$=1-0.3$

$=0.7$

${P }_{ A }={ P }_{ A }^{ 0 }{ X }_{ A }$

$=300\times 0.3$

$=90mm\quad of\quad Hg$

${P }_{ B }={ P }_{ B }^{ 0 }\times { X }_{ B }$

$=450\times 0.7$

$=315 mm\quad of\quad Hg$
Now in the vapour phase:
Mole fraction of liquid A,

$A=\dfrac { { P }_{ A } }{ \left( { P }_{ A }+{ P }_{ B } \right) } =\dfrac { 90 }{ 90+315 } =0.22$

Thus mole fraction of liquid A is 0.22

The vapor pressure of water is $12.3 \text { }KPa$ at $300K$ . Calculate the vapor pressure of $1$ molal solution of a non-volatile solute in it.

$P_{H_2O}=P_A=12.3 \ K \ Pascal$

Let moles of solute be $1$
and mass of water be $1000 \ g = 1 \ kg$

Number of moles of water $= \cfrac {1000}{18}=55.5 \ moles$

According to Raoult's law of
vapor pressure of solute

$\Rightarrow P_{solute}= \chi _{solute} \times P_{solute/total}$

$P_{solute}= \cfrac 1{56.5} \times 12300 \ Pascal$

$=217.7 \ Pascal$

What is vapour pressure?

The pressure exerted by vapours of liquid on the surface of the liquid when equilibrium is established between liquid and it's vapour is called vapour pressure of a liquid.

The vapour pressure of water is $92$ mm at $323$ K. $18.1$ g of urea are dissolved in $100$ g of water. The vapour pressure is reduced by $5$ mm. Calculate the molar mass of urea.

Given:

$P\overset {o}{_A}= 92 mm, P\overset {o}{_A}-P_A=5mm$

$W_A=100, W_B=18.1,$

$M_A=18$

$M_B=?$

$n_{A}=\cfrac {W_A}{M_A}= \cfrac {100}{18}=5.56$

$n_B=\cfrac {W_B}{M_B}=\cfrac {18.1}{M_B}$

$\Rightarrow \cfrac {P\overset {o}{_A}-P_{A}}{P\overset {o}{_A}}=\cfrac {n_B}{n_A+n_B}$

$\Rightarrow \cfrac {5}{92}=\cfrac {18.1/M_B}{5.56+\cfrac {18.1}{M_B}}$

$\Rightarrow 0.054\left(5.56+\cfrac {18.1}{M_B}\right)=\cfrac {18.1}{M_B}$

$\Rightarrow 0.3+\cfrac {0.97}{M_B}=\cfrac {18.1}{M_B}$

$\Rightarrow 0.3=\cfrac {17.73}{M_B}\Rightarrow M_B=\cfrac {17.73}{0.3}= 57.1$

$\therefore$ Molar mass of Urea=

$57.1 g/mol$

If equal volume of ${ BaCl }_{ 2 }$ and $NaF$ solutions are mixed , which of these combination will give a precipitate?
${ K }_{ sp }$ of ${ BaF}_{ 2} ={ 1.7\times 10 }^{ -7 }.$

If equal of

$BaCI2(aq)$ and

$NaF(aq)$ solution are mixed together, which will NOT give a precipitate

$(Ksp\ of\ BaF2=1.7\times 10-7)?$

this is correct question

a. all will produce a precipitate

$0.0040\ M\ BaCl2$ and

$0.020\ M\ NaF$

$0.010\ M\ BaCl2$ and

$0.0020\ M\ Naf$

$0.020\ M= BaCl2$ and

$0.0020\ M\ NaF$

$0.0150\ M\ BaCl2$ and

$0.010\ M\ NaF$

Only tune mixture will precipitate which have greater ionic product then

$Ksp$ value

$Ksp (BaF_{2})=[Ba+2][F^{-}]^{2}$

(b) ionic produced

$=0.0040\times (0.02)^{2}=1\times 10^{-7}$

it will give precipitate

$=0.010\times (0.015)^{2}=22.5\times 10^{-7}$

it will also give precipitate

(d) ionic product

$=0.02\times (0.002)^{2}=0.8\times 10^{-7}$

it will not give precipitate

(e) ionic product

$=0.016\times (0.010)^{2}$

$=15\times 10^{-7}$

it will also give precipitate

only option

$c$ will not give precipitate

The vapour pressure of a pure liquid A is $80$ mmHg at $300K$ . It forms an ideal solution with liquid B. When the mole fraction of B is $0.4$ , the total pressure was to be $88$ mmHg. The vapour pressure of liquid B would be?

We know that,

$H_AP_A+H_BP_B=P$

$\Rightarrow 0.6 \times 88+0.4P_B=88$

$\Rightarrow P_B=\cfrac {(88-0.6\times 80)}{0.4}$

$=100mm Hg$ .

The vapour pressure of solution of 5g of non electrolyte in 100 g of a water at a particular temperature is $2985 N{ m }^{ -2 }$ . The vapour pressure of pure water at that temperature is $3000 N{ m }^{ -2 }$

Mole fraction of solute,

$x_2=\cfrac {-\Delta p}{p_1^*}=\cfrac {3000-2985}{3000}=0.005$

Now,

$x_2=\cfrac {m_2/M_2}{\cfrac {m_1}{M_1}+\cfrac {m_2}{M_2}}\approx \cfrac {m_2/M_2}{m_1/M_1}$

$\therefore 0.005= \cfrac {5/M_2}{100/18}$

$\Rightarrow M_2=180 gmol^{-1}$

$\therefore$ The molar mass of solute is

$180 gmol^{-1}$ .

What is the mass of a non volatile solute (molecular mass 60) that needs tobe dissolved in $100 g$ of water In order to decrease the vapour pressure of water by $25%$ .What will be the velocity of solution?

Given that

$P^o=100, P=25, M=60, W=100, MH_2O=18$ .

$\Rightarrow \cfrac {P^o-P}{P^o}=\cfrac {W}{m}\times \cfrac {M}{W}$

$\Rightarrow \cfrac {100-25}{100}=\cfrac {W}{60}\times \cfrac {18}{100}$

$W=250 gm$

Molality=

$\cfrac {\text{Mole of solute}}{\text{Weight of solvent}}$

$=\cfrac {250\times 1000}{60 \times 1000}=41.66 m$ .

If p° and p are the vapour pressure of a solvent and it's solution repetitively and N1 and N2 are the mole fractions of the solvent and solute respectively, then the relation between them is?

1313658_1124235_ans_15d1c79c2116467790b2114b307ac89c.jpeg

The vapour pressure of pure water is 23.8 mm Hg at $25^0C$ What is the vapour pressure of 2.50 molal $C_6H_{12}O_6$ .

First convert the molality to a mole fraction

$\Rightarrow 2.50m$

$C_6H_{12}O_6=2.50$ mol/

$1kg$ of

$H_2O$

$\Rightarrow 1000g/18.015= 55.51$ mol

$\Rightarrow 55.51+2.50=58.01$ mol

Mole fraction of

$H_2O=\cfrac {55.81}{58.01}=0.9569$

Raoult's Law:

$P_{solution}=(Y solvent)(P^osolvent)$

$=(0.9569)(23.8)=22.8 mm Hg$ .

2 g benzoic acid $(C_{6}H_{5}COOH)$ dissolved in 25 g of benzene shoes a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg $mol^{-1}$ . What is the percentage association of acid if it forms dimer in solution?

Here solute is bengoic acid and soluent is bengelnt

$W$ solute

$=2g$

$M$ solute

$(6H_6)=6\times 12+6=78 g/mole$

$K_f=4.9\ K\ kg/mole$

$\Delta T_f=1.62\ K$

$2C_6H_5COOH\rightleftharpoons (6H_5COOH)_2$

$'x'$ is degree of association then we would have

$(1-x)$ mol of benxoic acid left undissociation.

Now total no. of moles of particles at equilibrium

$=1-x+ \dfrac {x}{2}=1-x---(1)$

$=i$

$i=\dfrac {Normal\ Molecular\ mass}{Normal\ Molecular\ mass}$

$=\dfrac {122}{241.98}----(2)$

$M(abnormal)=\dfrac {4.9\times 2\times 1000}{1.62\times 25}=241.98\ g/mol$

From

$(1)$ &

$(2)\ \dfrac {122}{241.98}=-\dfrac {x}{2}$

$\dfrac {x}{2}=\dfrac {241.98-122}{241.98}=99.2\%$

Degree of association of benzoic zcid

$=99.2\%$

Explain homogenous and heterogenous conditions with an example.

A homogeneous population is one in which the individuals in the population have little or no population. E.g: In a population residing in an island where there is no migration of individuals into the island, all the individuals will be homogenous for a particular trait such as disease immunity. A heterogenous population is one in which a large degree of variation is seen in traits. E.g: a population in which individuals have all migrated from different regions will be heterogeneous with regard to traits such as disease immunity.

Derive van"t Hoff equation for osmotic pressure of a solution.

1350738_1223648_ans_177df31a97874d0e97ca84e7d87430a2.jpg

A sample of $O_{2}$ gas is collected over at $23^{o}\ C$ at a barometric pressure of $751\ mm\ Hg$ (vapour pressure of water at $23^{o}\ C$ is $21\ mmHg)$ . The partial pressure of $O_{2}$ gas in the sample collected is

1309083_1288890_ans_3b82c4a1dcfa4b969292a9728f0eba3f.jpg

A $0.2\%$ aq. sol. of a non-volatile solution exerted a VP of $1.004$ bar at $100^\circ C$ . What is the molar mass of the solute ?
(Given V.P of mass water at $100^\circ C$ is $1.013$ bar and molar mass of water is $18\,g/mol$ ).

1309070_1290256_ans_19e2f995c14b4512b43ee4276614a673.jpg

A 821 mL $N_2(g)$ was collected over liquid water at 300 K and 1 atm. If vapour pressure of $H_2O$ is 30 torr then moles of $N_2(g)$ in moist gas mixture is :

1309028_1283642_ans_d87f3a6501a647e5819fdfbf1f97b08f.jpg

Define Van't Hoff's factor.

1352551_1288983_ans_a0b0e201b0fd4167bc3ec68211f6d529.jpg

The vapour of a sodium containing $13\times 10^{-3}\ kg$ of solute in $0.1\ kg$ of water at $298\ K$ is $27.371\ mm\ Hg$ . Calculate the molar mass of the solute. Given that the vapour pressure of water at $298\ K$ is $28.065\ mm\ Hg$ .

1308782_1327894_ans_f40e09498f234f0f9717d83925883d0f.jpg

Find its mass of urea dissolved in $100g H_2O$ which decreese the vapour presure of solvent by $20\%$

1309067_1304107_ans_42d250792a7e4e79a0c7688dcb6267aa.jpg

110g of salt is present in 550g of solution. Calculate the mass percentage of the solution.

Mass percentage = mass of the solute / mass of the solution * 100

$\Rightarrow \dfrac{ 110}{550} \times 100$

$\Rightarrow \dfrac{ 1}{5} \times 100$

$\Rightarrow 20$ %

This, mass percentage of the solution is

$20$ %

The vapour pressure of pure water at is 25.21 torr. What is the vapour pressure of a solution which contain 20.0 glucose, ${C_6}{H_{{\text{12}}}}{{\text{O}}_6}$ in 70 g water ?

1308788_1416352_ans_52937fc25b2f4735932514de858866b3.jpg

Two A and B in a mixture produced 550 mm of Hg vapour pressure. If their vapour pressure in the pure states are 400 mm of Hg and 600 mm of Hg respectively. What is the vapour pressure of 'A' in the mixture?

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Calculate the percentage by mass to volume for 0.2 M NaOH solution.

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What is the number of molecules in 1.5 moles of ammonia ?

$\textbf{Given data:}$

Number of moles of ammonia = 1.5 moles

$\textbf{Step 1: Calculate the number of molecules in 1.5 moles of ammonia.}$

$Number\ of\ moles\ of\ ammonia = \dfrac{Number\ of\ molecules\ of ammonia}{Avogadro's\ number}$

$1.5 = \dfrac{Number\ of\ molecules\ of\ ammonia}{6.022 \times 10^{23}}$

$Number\ of\ molecules\ of\ ammonia = 1.5 \times 6.022 \times 10^{23}$

$Number\ of\ molecules\ of\ ammonia = 9.033\times 10^{23}$

$\textbf{Conclusion:}$

Therefore, the number of molecules in 1.5 moles of ammonia is $9.033 \times 10^{23}$ .

When $HgI_2$ is added to aqueous solution of KI, why there is an increase in vapour pressure of solution ?

$KI$ forms a complex with

$Hg{I}_{2}$ .

$2 KI + Hg{I}_{2} \longrightarrow {K}_{2}Hg{I}_{4}$

Therefore, the number of particles in solution decreases resulting in the decrease in relative lowering of vapour pressure. Hence, vapour pressure increases.

Under what condition the van't Hoff factor is greater than one?

In case of dissociation of strong electrolyte, van't Hoff's factor is more than one as the strong electrolyte will form more than

$1$ ion on dissociation.

100 g of liquid A (mol. mass of = 140 g/mol) was dissolved in 1000 g of liquid B (molar mass of = 180 g/mol). The vapour pressure of pure liquid B was found to be 500 atmosphere. Calculate the vapour pressure of liquid A and its vapour pressure in the solution, if the total vapour pressure of the solution is 475 atmosphere.

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Predict whether van't HOff factor $(i)$ is less than one or greater than one in the following:
(i) $CH_3COOH$ dissolved in water
(ii) $CH_3COOH$ dissolved in benzene

(i) When acetic acid dissolves in water, it dissociates into $C{H}_{3}COO^-$ and ${H}^{+}$ ions and thus the Van't Hoff factor is greater than $1$ .
(ii) Acetic acid dimerizes when dissolved in benzene and thus the Van't Hoff factor is less than $1$ .

The vapour pressure of water at $20^oC$ is $17$ mm Hg.Calculate the vapour pressure of a solution containing $2.8$ g of urea $(NH_2CONH_2)$ in $50$ g of water ( $16.71$ mm Hg).

1308791_1562989_ans_4043279ad43b4cf0946145816d923445.jpg

Octane is a component of gasoline. Complete combustion of octane leads to $CO_2$ and $H_2O$ while incomplete combustion produces $CO$ and $H_2O$ , which not only reduced the efficiency of the engine using the fuel but is also toxic.In a certain test run, gallon of octane is burned in an engine, The total mass of $CO, CO_2$ and $H_2O$ produced is $9.768\ kg$ . Calculate the efficiency of the process, i.e, calculate the percentage of octane converted to $CO_2$ . The density of octane is $2.28\ kg/gallon$ .

Octane :-

$C_8H_{18}$ ;

$M_{c8}H_{18}=114\ gm.$

$1$ gallon

$=2.28kg=2280\ gm$ octane

$\therefore n_{octane}=\dfrac{2280}{114}=20$ mole

$C_8H_{18} + \dfrac{17}{2}O_2 \longrightarrow 8_{CO}+9H_2O$

$20-x$

$8(20-x)$

$9(20-x)$

$C_8H_{18} + \dfrac{25}{2}O_2 \longrightarrow 8CO_2+9H_2O$

$x$

$8x$

$9x$
Let octane complete combustion forming

$CO_2$ is

$x$ mole

$\therefore$ octane forming

$CO$ is

$20-x$ mole

$\therefore$ Total weight of

$CO,CO_2,H_2O=9.768kg=9768g$

$8_x \times 44+8(20-n)\times 28+\left\{9(20-x)+9x\right\} \times 18=9768$

$\Rightarrow 352x+160\ k\ 28-224x+180\times 18=9768$

$\Rightarrow 352x-224x+4480+3240=9768$

$\Rightarrow 128x=2048$

$x=16$

$\therefore$ Octane converted to

$CO_2=16\ mole$

$\therefore\ \%$ of a octane converted to

$CO_2=\dfrac{16}{20}\times 100=80y.$

$\therefore\ \%$ efficiency is

$80\%$ as only

$80\%$ octane converted to

$CO_2$

$\therefore$ is

$0080$

In $1.684\ g$ sample of a mixture of $MgSO_{4}.7H_{2}O$ and $MgCl_{2}.6H_{2}O$ containg some inert impurity was subjected to suitable treatment, as a result of which there were obtained $0.699\ g$ of $BaSO_{4}$ adn $0.888\ g$ of $Mg_{2}P_{2}O_{7}$ . The mass percentage of impurity is $(Ba=137, Mg=24, P=31)$

Mole of

$BaSO_4=\dfrac{0.699}{233.38}=3\times 10^{-3}mole$

$\therefore$ Mole of

$MgSO_4. 7H_{20}=3\times 10^{-3}mole$
Man of

$MgSO_4.7H_{20}=3\times 10_{-3}\times 246$

$=0.738\ g$
Total mole of

$Mg_2P_2O_4=\dfrac{0.888}{222}=4\times 10^{-3}mole$
Mole of

$MgCl_2.6H_{20}=4\times 10^{-3}$
Mass of

$MgCl_2.6H_{20}=4\times 10^{-3}\times 203$

$=0.812\ g$
Man of Impurity

$=1.684=(0.738+0.812)$

$1.684-1.55$

$=0.134 g$

$\%$ Impurity

$=\dfrac{0.134}{1.684}\times 100$

$=7.95\%$
Answer is

$8\%$

A $1.174\ g$ sample of special grade steel was treated approximately with Chugaev's reagent by which nickel was precipitated as nickel dimethylglyoxime $NiC_{8}H_{14}O_{4}N_{4}$ . The dried precipitate weighed $0.2136\ g$
The percentage of nickel in the steel being analysed is $(Ni=58.7)$

$Ni^{2+}+2\ dmg\rightarrow NiC_{8}H_{14}O_{4}N_{4}$

$58.7$
Molar mass of

$NiC_{8}H_{14}O_{4}N_{4}$

$=58.7+8\times 12+14\times 1+4\times 16+4\times 14$

$=288.69$ gram/mole

$\therefore$ From balanced equation

$58.7$ gram

$Ni^{2+}\equiv 288.69$ gram

$NiC_{8}H_{14}O_{4}N_{4}$

$\therefore x$ gram

$Ni^{2+}\equiv 0.2136$ gram

$x=\dfrac{58.7\times 0.2136}{288.69}$

$=0.0434$ gram

$Ni^{2+}$

$\%$ og

$Ni$ in sample

$=\dfrac{0.0434}{1.174}\times 100$

$=3.69\%$

A blackened silver coin weighing $15\ g$ on treatment with $HCl$ yielded $25\ mL$ of $H_{2}S$ at $12^{o}C$ and $775\ mm$ pressure. What percentage of original silver tarnished?

$1.57\%$

$3\ g$ sample of blue vitrol $(CuSO_{4}.5H_{2}O)$ were dissolved in water. $BaCl_{2}$ solution was mixed in excess to this solution. The precipitate obtained was washed and dried. It weighed $2.82\ g$ . Determine $\%$ of $SO_{4}^{2-}$ by weight in sample.

$38.72\%$

What is the percentage by volume of gases in the atmosphere?

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An aqueous solution containing $20\%$ by weight of liquid $X(mol. wt. =140)$ has a vapour pressure $160\ mm$ at $60^{o}C$ . Calculate the vapour pressure of pure liquid $X$ if the vapour pressure of water is $150\ mm$ at $60^{o}C$

$X_{Solute}=\dfrac{20/140}{\dfrac{20}{140}+\dfrac{80}{18}}=\dfrac{0.1428}{4.587}=0.03$

$P_{solution}=X_{Solute}.P_{Solute}+X_{solvent}+P_{Solvent}$

$160=0.031\times P^{0}solute+(1-0.0311)\times 150$

$160-144.42=0.03.P^{0}_{solute}$

$\dfrac{14.58}{0.031}=P^{0}_{Solute}$

$P^{0}_{Solute}=470.5\ mm$ of

$Hg$

Which of the following aqueous solutions has a higher vapour pressure if the density of water is $1\ g/cc$ ?
(1) Solution having mole fraction of cane sugar $=0.1$
(2) Solution having molal concentration = 1 m
[Give your answer in terms of 1 or 2, e.g. enter 1 if (1) is true.]

$X_{Solute}=\dfrac{P_{water}-P_{sol}}{P_{Water}}\Rightarrow P_{sol}=P_{water}(1-X_{Solute})$

For molality

$=1\ m;X_{Solute}=\dfrac{n_{Solute}}{n_{Solute}+n_{Solvent}}=\dfrac{\dfrac{n_{Solute}}{wt_{Solvent}}}{\dfrac{n_{solute}}{wt_{Solvent}}+\dfrac{n_{Solvent}}{wt_{Solvent}}}$

$X_{solute}=\dfrac{\dfrac{m}{1000}}{\dfrac{m}{1000}+\dfrac{1}{18}}=\dfrac{\dfrac{1}{1000}}{\dfrac{1}{1000}+\dfrac{1}{18}}=\dfrac{0.001}{0.001+0.055}=0.0177$

$\therefore P_{solution}=P_{water}(1-x_{solute})$

For

$(b), X_{Solute}$ is less, so

$P_{solution}$ will be higher.

Ans-

$2$

Which of the following aqueous solutions has a higher vapour pressure if the density of water is $1\ g/cc$ ?
Solution having molal concentration $=1\ m$

$X_{Solute}=\dfrac{P_{water}-P_{sol}}{P_{Water}}\Rightarrow P_{sol}=P_{water}(1-X_{Solute})$

For molality

$=1\ m;X_{Solute}=\dfrac{n_{Solute}}{n_{Solute}+n_{Solvent}}=\dfrac{\dfrac{n_{Solute}}{wt_{Solvent}}}{\dfrac{n_{solute}}{wt_{Solvent}}+\dfrac{n_{Solvent}}{wt_{Solvent}}}$

$X_{solute}=\dfrac{\dfrac{m}{1000}}{\dfrac{m}{1000}+\dfrac{1}{18}}=\dfrac{\dfrac{1}{1000}}{\dfrac{1}{1000}+\dfrac{1}{18}}=\dfrac{0.001}{0.001+0.055}=0.0177$

$\therefore P_{solution}=P_{water}(1-x_{solute})$

For

$(b), X_{Solute}$ is less, so

$P_{solution}$ will be higher.

What is the vapour pressure at $100^{o}C$ of a solution containing $15.6\ g$ of water and $1.68\ g$ of sucrose $(C_{12}H_{22}O_{11})$ ?

$X_{Sucorse}=\dfrac{\dfrac{1.68}{342}}{\dfrac{1.68}{342}+\dfrac{15.6}{18}}=\dfrac{\dfrac{1}{200}}{\dfrac{1}{200}+0.867}=\dfrac{5\times 10^{-3}}{0.872}=5.734\times 10^{-3}$

$100^{o}C, P^{0}_{Solvent}=760\ torr$

$\therefore \dfrac{P^{0}_{Solvent}-P_{Solution}}{P^{0}_{Solvent}}=5.734\times 10^{-3}$

$P_{solution}=760-760\times 5.734\times 10^{-3}$

$=760-4.358=755.6\ torr$

$\therefore P_{Solution}=755.6\ torr=755.56\ cm$ of

$Hg$

$(\because 1\ torr=1\ mm\ of\ Hg)$

Find the percentage of nitrogen in an organic compound analysed by Kjeldahl method. $1.61 \,g$ of the compound produced $NH_3$ which was absorbed in $250 \,mL$ of $\dfrac{N}{2}H_2SO_4$ solution.The remaining acid was then diluted to one litre, $25 \,mL$ of which required $25.5 \,mL$ of $\dfrac{N}{10} NaOH$ for exact neutralization.

$25mL \rightarrow 25.5 mL\times \cfrac{1}{10}$

$1000 mL \rightarrow \cfrac{25.5}{25} \times \cfrac{1}{10} \times 1000$

$=102\ mL\ N$ NaOH

$= 204\ mL$ of

$\cfrac{N}{2}$ NaOH

Acid used for neutralisation of

$NH_3= 250-204=46\ mL$

$\%$ of Nitrogen in the given organic compound by Kjeldahl method

$=1.4 \times N_1 \times \cfrac{V}{W}$

Given:

$N_1 = \cfrac{1}{2}, W= 1.61\ g$
and

$V=46\ mL$

$\%\ of\ N= 20\%$

Ans -

$20$

One litre of water was added to $500\ mL$ of $32\%\ HNO_{3}$ of density $1.20\ g/mL$ . What is the per cent concentration of $HNO_{3}$ in the solution obtained?

$V_{Water}=1\ lit, d=1000\ g/lit$

$W_{Water}=V.d=1000\ g$

wt of

$sol^{n}=500\times 1.20=600\ g$

Total wt of sol

$=1600\ g$

Also wt of

$HNO_{3}=600\times \dfrac{32}{100}192\ g$

$\therefore \%$ by wt of

$HNO_{3}=\dfrac{192}{1600}\times 100=12.0\%$

At $25^{o}C$ benzene and toluene have densities $0.879$ and $0.867\ g/cc$ respectively.
Assuming that benzene-toluene solutions are ideal. establish the equation for the density of the solution.
$d=\dfrac{1}{100}\left\{0.879\ V+0.867 (100-V)\right\}$ where $V$ is the volume per cent of benzene.

$d_{benzene}=0.879\ g/cc$ &

$d_{toulene}=0.867\ g/cc$

$V$ is the volume

$\%$ of benzene

$\%$ of toluene

$=(100-V)\ ml$

$\therefore$ Total mass

$(100\ ml)=V_{benzene}\times d_{benzene}+V_{total}\times d_{toluene}$

$=V\times 0.879+(100-V)\times 0.867$

$\therefore d_{sol}=\dfrac{total\ mass}{total\ vol}=\dfrac{0.879\ V+0.867(100-V)}{100}$

$0.5\ g$ of a nonvolatile organic compound $(mol.wt. 65)$ is dissolved in $100\ mL$ of $CCl_{4}$ . If the vapour pressure pure $CCl_{4}$ is $143\ mm$ , what would be the vapour pressure of the solution? (Density of $CCl_{4}$ solution is $1.58\ g/cc$ )

$P^{0}_{CCl_{4}}=143\ mm$ of

$Hg$

Wt of

$CCl_{4}=V_{CCl_{4}}\times d_{ccl_{4}}=100\times 1.58=158\ g$

$\therefore n_{CCl_{4}}=\dfrac{158}{12+4\times 35.5}=\dfrac{158}{154}=1.026$

$W_{non\ volatile} =0.5\ g$

$n_{Solute}=\dfrac{0.5}{65}=7.7\times 10^{-3}$

$\therefore X_{Solute}=\dfrac{n_{Solute}}{n_{total}}=\dfrac{7.7\times 10^{-3}}{1.0337}=\dfrac{143-P_{solution}}{143}$

$P_{solution}=143-143\times 7.45\times 10^{-3}=141.93\ mm$ of

$Hg$

Determine the per cent concentration of a solution obtained by mixing $300\ g$ of a $25\%$ and $400\ g$ of a $40\%$ solution.

Total wt of solution

$=300+400=700\ g$

Total wt of solute

$=3\rho 0\times \dfrac{25}{100}+400\times \dfrac{40}{100}$

$=75+160=235\ g$

$\therefore \%$ by wt (Now)

$=\dfrac{235\ g}{700\ g}\times 100=33.57\%$

When $10$ mL of ethanol of density $0.7893$ g/L is mixed with $20$ mL of water of density $0.9971$ g/L at $25^o$ C, the final solution has a density of $0.9571$ g/L. Calculate the percentage change in total volume on mixing.

$Mass = density \times volume$

Mass of ethyl alcohol =

$0.7893 \times 10 =7.893 \ g$

Mass of water =

$0.9971 \times 20 = 19.942 \ g$

Total mass of the solution = 7.893 + 19.942 = 27.835 g

Volume of solution after mixing =

$\dfrac{Mass}{Density} = \dfrac{27.835}{0.9571} = 29.083 \ ml$

Total volume of ethanol and water before mixing = 10 +20 = 30 ml

The percentage change in total volume on mixing is =

$\dfrac{30 - 29.083}{30} \times = 3.05 \%$

A solution contains $1$ milli-curie of L-phenyl alanine $C^{14}$ (uniformly labelled) in $2.0\ mL$ solution. The activity of the labelled sample is given as $150$ milli-curie/milli-mole. Calculate:
(a) The concentration of the sample in the solution in mole/litre.
(b) The activity of the solution in terms of counting per minute/mL at a counting efficiency of $80\%$ .

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Calculate the concentration of a solution obtained by mixing $300\,g \,25$ % by weight solution of $NH_4Cl$ and $150\,g$ of $40$ % by weight solution of $NH_4Cl.$

Weight of solution I and II

$= 300 + 150 = 450\,g$

Weight of

$NH_4Cl$ in I solution

$= \dfrac {25 \times 300}{100} = 75\,g$

Weight of

$NH_4Cl$ in II solution

$= \dfrac {40 \times 150}{100} = 60\,g$

$\therefore$ Total weight of

$NH_4Cl = 75 + 60 = 135\,g$

$\therefore \%$ by wt. of mixed solution

$= \dfrac {135}{450} \times 100 = 30\%$

If the vapour pressure of pure liquids A and B are $300$ mm and $800$ mm Hg at $75^o$ C, calculate the composition of the mixture such that it boils at $75^o$ C. Find the composition of the vapour phase.

1702765_1738268_ans_15f955bb7ed94bf08f0496316b6c94e7.jpg

Calculate the % of free $SO_3$ in oleum (a solution of $SO_3$ in $H_2SO_4$ ) that is labelled $109$ % $H_2SO_4$ by weight.

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A $0.5$ g sample containing $MnO_2$ is treated with HCl, liberating $Cl_2$ . The $Cl_2$ is passed into a solution of KI and $30.0 cm^3$ of $0.1$ M $Na_2S_2O_3$ are required to titrate the liberated iodine. Calculate the percentage of $MnO_2$ in the sample.

1740786_1739120_ans_e11bc6dc9bf3490e9c27c1ac7bbd6e52.jpg

What type of liquids form ideal solutions?

Liquids that have similar structures and polarities form ideal solutions.

Explain the solubility rule "like dissolve like" in terms of intermolecular forces that exist in solutions.

A substance (solute) dissolves in a solvent if the intermolecular interactions are similar in both the components; for example, polar solutes dissolves in polar solvents and non polar solutes in non polar solvents thus we can say "like dissolves like".

State any two characteristics of ideal solutions.

Ideal solutions (i) obey Raoult's law (ii)

$\Delta_{mix} H = 0 , \Delta_{mix} V = 0$

Neutralisation of $30 \,gm$ of a mixture of acetic acid and phenol solutions required $100 \,ml$ of $2M$ sodium hydroxide solution. When the same mixture was treated with bromine water, $33.1 \,gm$ of precipitate was formed. Determine the mass percentage of acetic acid and phenol in the given solution.

$331 \,gm$ of

$(B)$ is obtained from

$94 \,gm$ of

$(A)$ .

ii.

$331 \,gm \,(B)$ is obtained from

$= \dfrac{94}{331} \times 33.1$

$= 9.4 \,gm$ of phenol
Weight of phenol

$= 9.4 \,gm$

$= \dfrac{9.4}{94} = 0.1 \,mol$

ii.

$NaOH$ will react with both

$CH_3COOH$ and phenol. Total molar equivalent of

$NaOH = 100 \times 2$

$= 200 \,mEq$

$= \dfrac{200}{1000} = 0.2 \,Eq$ of

$NaOH$

$= 0.2 \,mol$ of

$NaOH$
Acid

$+$ Phenol

$= 2.0 \,mol$
Acid

$+ \,0.1 \,mol = 2.0 \,mol$

$\therefore$ Acid

$= 0.2 - 0.1 = 0.1 \,mol$

$= 0.1 \,Eq.$

$1 \,Eq.$ of

$CH_3COOH = 60 \,gm$

$0.1 \,Eq$ of

$CH_3COOH = 6 \,gm$
Weight of acid

$= 6 \,gm$

Mass percentage of acid

$= \dfrac{6}{30} \times 100 = 20 \%$

Mass percentage of phenol

$= \dfrac{9.4}{30} \times 100 = 31.3 \%$

Define w/w (percentage mass by mass) way of expressing concentrations.

Mass percentage (w/w) : It is weight by weight relationship. Thus mathematically,
Mass % of a component

$= \frac{W \space of \space the \space component \space in \space the \space solution} {W \space of \space the \space solution} \times 100$
where, W represents weight of the component or solution

Both Sarika and Mohan were asked to make a salt solution. Sarika was given a teaspoonful of salt and half a glass of water, whereas Mohan was given twenty teaspoons full of salt and half a glass of water.
(a) How would they make salt solutions?
(b) Who would be able to prepare a saturated solution?

a) Sarika and Mohan both will mix the salt with water by stirring it continuously. A solution will be formed.
b) As Mohan has more salt, his solution will be in a saturation state. Some salt will not be dissolved

Match the terms given in Column I with expressions given in Column II.

(A) - (4)

(B) - (3)

(D) - (5)

(E) - (1)

Match the items given in Column I with Column II.

$\text { (A) - (5), } \text { (B) - (3), } \text { (C) - (4), } \text { (D) - (2), } \text { (E) - (1) }$

Why is the vapour pressure of an aqueous solution of glucose lower than that of water?

Vapour pressure of any solvent or water is caused due to escaping tendencies of the water molecules from the liquid level/surface. In pure water, only water molecules are present at its surface, but when a non volatile solute like glucose is dissolved in it, certain number of nonvolatile glucose molecules, with no escaping tendency are also present at the surface of aqueous solutions.

The number of water molecules at the surface is correspondingly decreased, due to which relatively lesser number of molecules of water can escape out as vapours. This results in lowering * of vapour pressure of water in its glucose solution as compared to that of pure water/solvent.

Such a relative lowering of vapour pressure is termed as its colligative property.

Define the v/v (volume by volume percentage) mode of expressing the concentration of a solution.

Volume % of a component

$= \frac{V \space of \space the \space component} {V \space of \space the \space solution} \times 100$
where, V represents volume.

Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature, however molarity is a function of temperature. Explain.

Molarity of solution is a weight by volume relationship to represent its strength and defined as 'the number of moles of solute dissolved in one litre of solution'. Since volume depends on temperature and undergoes a change with change in temperature, the molarity will also change with change in temperature.

On the other hand, the other concentration terms such as mass percentage, ppm, mole fraction and molarity are based upon mass by mass relationship of solute and solvent present in a binary solution. Mass does not change with change in temperature, as a result these concentration terms remain unchanged with variation of temperature.

According to the definition of all these terms, mass of the solvent used for making the solution is related to the mass of solute.

Will honey form a solution in water? (Yes/No)

YES

Honey is already a solution of sugars (and a few other solutes) in water, and mixing it with water produces a more dilute solution.

Define saturated solution

Saturated solution : When a solution cannot dissolve any more of solute at a given temperature, it is called saturated solution.

Differentiate between the following.
Solution and Suspension

Solution and Suspension :

Solution :

1. It is an example of homogeneous mixture.

2. It is formed when a solid dissolves in liquid.

3. For example - sugar dissolved in water.

Suspension :

1. It is an example of heterogeneous mixture.

2. It is formed when an insoluble solid is added to solvent.

3. For example - chalk dissolved in water.

Give an example of a reaction where the following are involved in a Solution

$NaCl(aq) +AgNO_3(aq) \rightarrow AgCl \downarrow_{white ppt} + NaNO_3(aq)$ is an example of a reaction where the following are involved in a Solution .

Define Seeding.

Seeding is a process in which a small number of crystals are used to produce more amount of crystals of the same material.

Explain why a solution is always clear and transparent.

Water-soluble substances dissolve completely in water and remain disappeared. After this water retains its property, hence solution remains clear and transparent in a solution.

Suggest a suitable technique to separate the constituents of the following mixtures. Also give the reason for selecting the particular method.
(a) Salt from sea water
(b) Ammonium chloride from sand
(c) Chalk powder from water
(d) Iron from sulphur
(e) Water and alcohol
(f) Sodium chloride and potassium nitrate
(g) Calcium carbonate and sodium chloride

(a) The technique used to separate the salt from seawater is Evaporation.

Reason - Because this method is used to separate the components of the homogeneous solid-liquid mixture. In this method sea water is collected in a shallow bed and allowed to evaporate in the sun. When all the water is evaporated, salt is left behind. By this method, we only get solid and liquid is evaporated In Its vapour form.

(b) Technique used to separate Ammonium chloride from sand is sublimation. Because this method is used for solid mixtures in which one of the components can sublime on heating. In this method. Ammonium chloride changes into vapours on heating and salt is left behind.

Reason - Because this process is used to separate the components of a heterogeneous solid-liquid mixture. In which solids are lights and insoluble In liquids. Substances used as filters are sand filter paper at C. These filters allows the liquid to pass through them but not solids.

(d) Technique to separate iron from sulphur is magnetic separation. Because. this method is used when one of the component of mixture is iron. Iron gets attracted towards the magnet and hence get separated.

(e) Technique used to separate water and alcohol is Fractional Distillation. Because in this method, the vapours of water is left behind In the original vessel as the alcohol boils at lower temperature than water. Thus these two liquids can be separated.

(f) Technique used is Fractional-crystallisation. Because this method is used when solubility of solid components of mixture and different in the same solvent. Here, sodium chloride and potassium nitrate both are soluble in water but solubility of potassium nitrate is more.

(g) Technique used Is Solvent Extraction Method. Because, by this method salts get dissolve in water while calcium carbonate being insoluble in water settles down in the container. And hence get separated about.

Will sodium carbonate form a solution in water? (Yes/No)

Yes

Sodium carbonate (washing soda) is a salt, hence forms solution in water.

What type of solutions are formed on dissolving different concentrations of soap in the water?

At lower concentration soap forms a normal electrolytic solution with water. After a certain concentration called critical micelle concentration, colloidal solution is formed.

Calculate the mass of $95\%$ pure $MnO_2$ to produce $35.5 \,g$ of $Cl_2$ as per the following reaction:
$MnO_2 + 4HCl \rightarrow {}MnCl_2 + Cl_2 + H_2O.$ (At. mass of $Mn = 55$ )

$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + H_2O$

$87\;g$

$71 \,g$

$71 g$ of

$Cl_2$ is produced by

$87 g$ of

$MnO_2$

$35.5 \,g$ of

$Cl_2$ is produced by

$\dfrac{87}{71} \times 35.5 = 43.49 \,g$ of

$MnO_2$

Since

$MnO_2$ is

$95\%$ pure i.e.,

$95 g$ of

$MnO_2$ is present in

$100 g$ of

$MnO_2$ sample.

$43.49g$ of pure

$MnO_2$ will be present in

$\dfrac{43.39 \times 100}{95} = 45.8g$ of

$MnO_2$ sample.

A solution of sodium hydroxide contains $4.8 \,g$ of the substance per $dm^3,$
What volume of water has to be added to one $dm^3$ of the solution in order to make it exactly decinormal ?

$\underset{strong \,solution}{V_1 \times N_1} = \underset{diluted \,solution}{V_2 \times N_2}$

Substituting we get,

$1000 \times 0.12 = V_2 \times 0.1$

$\therefore V_2 = \dfrac{1000 \times 0.12}{0.1} = 1200$

$\therefore$ Volume of water that should be added to

$1000 \,cm^3$ of the solution

$= 1200 - 1000 = 200 \,cm^3$

Define: Mass percentage.

Mass fraction multiplied by

$100$ gives mass percentage. E.g.: mass percentage of

$A$ for binary mixture.

$W_A \%=\dfrac{W_A}{W_A + W_B} \times 100$

Which one in each of the following pairs of liquids is more viscous: coconut oil, castor oil

castor oil

castor oil is more viscous as it is denser than coconut oil.

What do you mean by decinormal solution ?

A decinormal solution is the solution containing

$0.1 \,g$ equivalent

$(\dfrac{1}{10}th)$ of solute / substance dissolved in

$1 \,dm^3$ of solution.

Calculate the volume of concentrated nitric acid of normality $14$ required to prepared $1 \,dm^3$ of $\dfrac{N}{10}$ nitric acid.

$\underset{conc. acid}{V_1 \times N_1} = \underset{dil. acid}{V_2 \times N_2}$

$V_1 \times 14 = 1 \times 0.1$

$V_1 = 0.007 L$

Calculate the percentage of nitrogen in $NH_4.$ (Atomic mass of $N = 14, H = 1$ amu)

Percentage of Nitrogen in

$NH_3 = \dfrac{Atomic \,weight \,of \,nitrogen}{Molecular \,weight \,of \,NH_4} \times 100$

$= \dfrac{14}{17} \times 100 = \dfrac{1400}{1700} = 82.3\%$

$1 \cdot 5 \mathrm{g}$ of pyrolusite ore were treated with 10 g of Mohr's salt and dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ . Atter the reaction, the solution was diluted to $250 \mathrm{cm}^{3} .$ $50 cm ^{3}$ of diluted solution required $10 \mathrm{cm}^{3}$ of $0 \cdot 1 \mathrm{N} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution.
Find out percentage of pure $\mathrm{Mn} \mathrm{O}_{2}$ in pyrolusite.

Pure

$\mathrm{MnO}_{2}$ present in pyrolusite oxidises

$\mathrm{Fe}^{2+}$ of Mohr's salt

$\left(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot \mathrm{FeSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right)$ to

$\mathrm{Fe}^{3+}$ .
Unreacted

$\mathrm{Fe}^{2+}$ of Mohr's salt is determined by

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
Chemical equations are:

$\begin{array}{c}\mathrm{MnO}_{2}+2 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+} \longrightarrow \\\mathrm{Mn}^{2+}+2 \mathrm{Fe}^{3+}+2 \mathrm{H}_{2} \mathrm{O} \\\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \longrightarrow \\2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\end{array}$
Step 1. To determine unreacted Mohr's salt.

$50 \mathrm{cm}^{3}$ of diluted Mohr's salt

$\equiv 10 \mathrm{cm}^{3} \text { of } 0 \cdot 1 \mathrm{N} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$

$\therefore \quad$ Normality of diluted solution

$=0.02 \mathrm{N}$ Mol. wt. of Mohr's salt

$=392$

$\therefore \quad$ Amount of unreacted Mohr's salt present in

$250 \mathrm{cm}^{3}$ solution

$=\dfrac{0 \cdot 02 \times 392}{4}=1 \cdot 96 g$
Amount of Mohr's salt used

$=10-1 \cdot 96=8 \cdot 04 \mathrm{g}$
From balanced equation,2 Moles of Mohr's salt (i.e.

$2 \times 392 \mathrm{g}$ ) react with one mole of

$\mathrm{MnO}_{2}=87 \mathrm{g}$

$\therefore \quad 8 \cdot 04 g$ of Mohr's salt will react with

$\mathrm{MnO}_{2}$

$=\dfrac{87}{2 \times 392} \times 8 \cdot 04 g=0 \cdot 892 g$
Now

$0 \cdot 892 \mathrm{g}$ of pure

$\mathrm{MnO}_{2}$ are present in

$1.5 \mathrm{g}$ of pyrolusite

$\begin{array}{l}\therefore \% \text { age of } \mathrm{MnO}_{2} \text { in pyrolusite }=\dfrac{0 \cdot 0892}{1 \cdot 5} \times 100 \\=59 \cdot 48 \%\end{array}$

$1 \cdot 1$ g of a sample of copper ore is dissolved and $\mathrm{Cu}^{2+}(a q)$ is treated with KI. The iodinethus liberated required $12 \cdot 12 \mathrm{cm}^{3}$ of $0.1 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ solution for titration. What is the percentage of copper in the ore?

The complete balanced equation for the redox reactions is

$2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-}+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{S}_{2} \mathrm{O}_{6}^{2-}+2 \mathrm{I}^{-}$
No. of moles of

$\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ used

$=\dfrac{12 \cdot 12}{1000} \times 0 \cdot 1$

$=1 \cdot 212 \times 10^{-3} \mathrm{moles}$
From the balanced equation,
2 moles of

$S_{2} \mathrm{O}_{3}^{2-}$ reduce

$\mathrm{Cu}^{2+}=2$ moles

$\therefore 1 \cdot 212 \times 10^{-3}$ moles of

$\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ will reduce

$\mathrm{Cu}^{2+}=1 \cdot 212 \times 10^{-3} \mathrm{moles}$

$\therefore \quad \mathrm{W} t .$ of pure Cu present in the ore

$=1 \cdot 212 \times 10^{-3} \times 63 \cdot 5=0 \cdot 077 g$
Thus, \%age of Cu in the ore

$=\dfrac{0 \cdot 077}{1 \cdot 1} \times 100$

$=7 \%$

What are the factors affecting the solubility?

Solubility is the maximum amount of a substance that will dissolve in a given amount of solvent at a specific temperature. There are two direct factors that affect solubility: temperature and pressure. Temperature affects the solubility of both solids and gases, but pressure only affects the solubility of gases

$5 \cdot 0$ g of a sample of brass were dissolved in 1 litre dil. $\mathbf{H}_{2} \mathbf{S O}_{4} \cdot \mathbf{2 0} \operatorname{cm}^{3}$ of this solution weremixed with KI and liberated lodine required $20 \mathrm{cm}^{3}$ of $0 \cdot 0327 \mathrm{M}$ hypo solution for titration. Calculate the percentage of copper in the alloy.

Brass is an alloy of Cu and Zn. When brass is treated with dil.

$\mathrm{H}_{2} \mathrm{SO}_{4}$ in presence of air,

$\mathrm{CuSO}_{4}$ and

$\mathrm{ZnSO}_{4}$ are obtained according to the following equations:

$2 \mathrm{Cu}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{CuSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}$

$\mathrm{Zn}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{ZnSO}_{4}+\mathrm{H}_{2}$
Out of

$\mathrm{CuSO}_{4}$ and

$\mathrm{ZnSO}_{4}$ , only

$\mathrm{CuSO}_{4}$ , reacts with KI to form

$I_{2}$ which can be titrated against hypo solution. The complete balanced equation for the redax reactions is

$\begin{array}{c}2 \mathrm{CuSO}_{4}+4 \mathrm{KI} \rightarrow 2 \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{CuI}_{2} \\2 \mathrm{CuI}_{2} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2} \\\mathrm{I}_{2}+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI} \\2 \mathrm{CuSO}_{4}+4 \mathrm{KI}+2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \longrightarrow \\\mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI}\end{array}$
Step 1. To find out the molariy of

$CuSO_4$ solution
Let the molarity of

$\mathrm{CuSO}_{4}$ soln.

$-\mathrm{M}_{1}$ Applying molarity equation,

$\dfrac{M_{1} V_{1}}{2}\left(\operatorname{CuSO}_{4}\right)=\dfrac{M_{2} V_{2}}{2}\left(N a_{2} S_{2} O_{3}\right)$
or

$M_{1} \times 20=0 \cdot 0327 \times 20$ or

$M_{1}=0 \cdot 0327$
Step 2 . To find out the percentage of copper in the alloy Volume of alloy solution

$=1000 \mathrm{cm}^{3}$
Molarity of alloy solution w.r.t.

$C u^{2+}$

$=0.0327 \mathrm{M}$
But At. wt. of

$C u=63.5$

$\therefore$ Amount of

$\mathrm{Cu}^{2+}$ formed

$=0.0327 \times 63 \cdot 5=2.0768$
But the amount of

$\mathrm{Cu}^{2+}$ ions in solution is equal to the amount of Cu in the alloy.

$\therefore$ Amount of copper in the alloy

$=2 \cdot 076 g$
But the amount of alloy (brass) taken

$=5 \cdot 0 \mathrm{g}$

$\therefore \quad \%$ of copper in the alloy

$=\dfrac{2 \cdot 076}{5} \times 100=41 \cdot 52 \%$
Thus, the percentage of copper in the alloy

$=41 \cdot 52 \%$

$1 \cdot 5 \mathrm{g}$ of pyrolusite ore were treated with 10 g of Mohr's salt and dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ . Atter the reaction, the solution was diluted to $250 \mathrm{cm}^{3} .$ $50 cm ^{3}$ of diluted solution required $10 \mathrm{cm}^{3}$ of $0 \cdot 1 \mathrm{N} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution.
Find out percentage of pure $\mathrm{Mn} \mathrm{O}_{2}$ in pyrolusite.

Pure

$\mathrm{MnO}_{2}$ present in pyrolusite oxidises

$\mathrm{Fe}^{2+}$ of Mohr's salt

$\left(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot \mathrm{FeSO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right)$ to

$\mathrm{Fe}^{3+}$ .
Unreacted

$\mathrm{Fe}^{2+}$ of Mohr's salt is determined by

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$
Chemical equations are:

$50 \mathrm{cm}^{3}$ of diluted Mohr's salt

$\equiv 10 \mathrm{cm}^{3} \text { of } 0 \cdot 1 \mathrm{N} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$

$\therefore \quad$ Normality of diluted solution

$=0.02 \mathrm{N}$ Mol. wt. of Mohr's salt

$=392$

$\therefore \quad$ Amount of unreacted Mohr's salt present in

$250 \mathrm{cm}^{3}$ solution

$=\dfrac{0 \cdot 02 \times 392}{4}=1 \cdot 96 g$
Amount of Mohr's salt used

$=10-1 \cdot 96=8 \cdot 04 \mathrm{g}$
From balanced equation,2 Moles of Mohr's salt (i.e.

$2 \times 392 \mathrm{g}$ ) react with one mole of

$\mathrm{MnO}_{2}=87 \mathrm{g}$

$\therefore \quad 8 \cdot 04 g$ of Mohr's salt will react with

$\mathrm{MnO}_{2}$

$=\dfrac{87}{2 \times 392} \times 8 \cdot 04 g=0 \cdot 892 g$
Now

$0 \cdot 892 \mathrm{g}$ of pure

$\mathrm{MnO}_{2}$ are present in

$1.5 \mathrm{g}$ of pyrolusite

$\begin{array}{l}\therefore \% \text { age of } \mathrm{MnO}_{2} \text { in pyrolusite }=\dfrac{0 \cdot 0892}{1 \cdot 5} \times 100 \\=59 \cdot 48 \%\end{array}$

Calculate the mass percentage of different elements present in sodium sulphate.

Molecular mass of sodium sulphate

$(Na_{2}SO_{4})$ is 142 g mole.

The elements present in sodium sulphate are sodium, sulphur, and oxygen.

Mass of sodium :

$23 \times 2 = 46 g$
Mass of sulphur :

$32\times1= 32 g$
Mass of oxygen :

$16 \times 4 = 64 g$

Mass percentage of sodium =

$\dfrac{32}{142} \times 100$ = 32. 38 %

$\approx$ 32.4 %

Mass percentage of oxygen =

$\dfrac{64}{142}\times100$ = 45.05 %

Mass percentage of sulphur = (100 - 32.4 - 45.05) % = 22.55 %

In an ore, the only oxidisable material is $S n^{2+}$ . This ore is titrated with a dichromate solution containing $2 \cdot 58$ of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ in 0.50 litre. The $0.40 \mathrm{g}$ sample of the ore required $10.0 \mathrm{cm}^{3}$ of titrant to reach equivalent point. Calculate the percentage of the in the ore. $(K=39 \cdot 1, C r=52,$ $S n=118 \cdot 7)$

Wt. of

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ present in

$500 \mathrm{cm}^{3}=2 \cdot 5 \mathrm{g}$

$\therefore \quad$ Wh. of

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ present in

$10 \mathrm{cm}^{3}$

$=\dfrac{2 \cdot 5}{500} \times 10 g$
Mot. wh. of

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=294$

$\therefore$ No. of moles of

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ present in

$10 \mathrm{cm}^{3}$
solution

$=\dfrac{2 \cdot 5 \times 10}{500 \times 294}=0.00017$
The balanced chemical equation for the redox reaction is :

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+3 \mathrm{Sn}^{2+} \rightarrow$

$2 \mathrm{Cr}^{3+}+3 \mathrm{Sn}^{4+}+7 \mathrm{H}_{2} \mathrm{O}$
From the above balanced equation,

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \equiv \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \equiv 3 \mathrm{Sn}^{2+}$
i. e., 1 mole

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ oxidises

$\mathrm{Sn}^{2+}=3$ moles

$\therefore 0.00017$ mole

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ will oxidise

$\mathrm{Sn}^{2+}$

$=3 \times 0.00017=0.00051$ mole
Amount of

$\mathrm{Sn}^{2+}$ oxidised

$=118 \cdot 7 \times 0 \cdot 00051$

$=0 \cdot 06 g$

$\therefore \%$ age of Sn in the ore

$=\dfrac{0 \cdot 06}{0 \cdot 40} \times 100=15$

(i) $200 {cm}^{3}$ of aqueous solution of a protein contains $1.26 g$ of protein. The osmotic pressure of solution at $300 K$ is found to be $8.3 \times {10}^{-2}$ bar. Calculate the molar mass of protein. $\left(R = 0.083 l bar {K}^{-1} {mol}^{-1}\right)$
(ii) What is the significance of Van't Hoff factor?

$(\pi =\frac{w\times 1000}{M\times V}\times R\times T)$

$(M =\frac{w\times 1000}{\pi\times V}\times R\times T)$

$(=\frac{1.26\times 1000\times 0.083\times 300}{2.57\times 10^{-3}\times 200})$

$= 61038 g (mol^{–1})$

Van toff factor is used to determine the extent of association or dissociation of solute in the solvent.

$i=\frac{Normal molecular mass}{observed molecular mass}$

In case of association the i value is less than one

In case of dissociation the i value is more than one

In case there is no association or dissociation the value is equals to one.

Derive the relationship between relative lowering of vapour pressure and molar mass of non-volatile solute.

Consider a solution in which

$W_2g$ of solute of molar mass

$M_2$ is dissolved in

$W_1g$ of solvent of molar mass

$M_1$ .

Hence, number of moles of solvent,

$n_1 = \dfrac{W_1}{M_1}$ and,

Number of moles of solute,

$n_2 = \dfrac{W_2}{M_2}$

$\therefore$ Total number of moles

$n = n_1 + n_2$

Mole fraction of the solvent,

$x_1 = \dfrac{n_1}{n}$

Mole fraction of the solute,

$x_2 = \dfrac{n_2}{n}$

In case of a dilute solution, the concentration of number of moles of the solute is very low, i.e.,

$n_1 >> n_2$

$\therefore n_1 + n_2 \simeq n_1$

Now, the mole fraction of the solute

$x_2$ is given by,

$x_2 = \dfrac{n_2}{n_1 + n_2} \simeq \dfrac{n_2}{n_1}$

$\therefore x_2 = \dfrac{W_2 / M_2}{W_1/ M_1}$

$\therefore x_2 = \dfrac{W_2 \times M_1}{W_1 \times M_2}$

$P_1^0$ and P are the vapour pressure of pure solvent and a solution respectively, then relative lowering of vapour pressure is given by,

$P = \dfrac{P_1^0 - P}{P_1^0}$

By Raoult's law

$\dfrac{\Delta P}{P_1^0} = \dfrac{P_1^0 - P}{P_1^0} = x_2$

$\dfrac{P_1^0 - P}{P_1^0} = \dfrac{W_2 M_1}{W_1 M_2}$

$\therefore \dfrac{\Delta P}{P_1^0} = \dfrac{n_2}{n_1} = \dfrac{W_2/ M_2}{W_1 / M_1} = \dfrac{W_2 M_1}{W_1 M_2}$

Hence, by determining the vapour pressure of a pure solvent and a solution, the molar mass of a non-volatile solute can be measured.

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol

(i) Phenol is partially soluble in water.
(ii) Toluene is water insoluble.
(iii) Formic acid is water soluble.
(iv) Ethylene glycol is water soluble.
(v) Chloroform is water insoluble.
(vi) Pentanol is partially soluble in water.

Two liquids A and B form an ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of A and 3 mole of B is 550 mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in their pure states.

Let the vapour pressure of pure A be

$=p^0_A$ ; and the vapour pressure of pure B be

$=p_B^0$ .

Total vapour pressure of solution (1 mole A + 3 mole B)

$=X_A\cdot p^0_A+X_B\cdot p^0_B$ [

$X_A$ is mole fraction of A and

$X_B$ is mole fraction of B]

$550=\dfrac{1}{4}p^0_A+\dfrac{3}{4}p^0_B$

$2200=p^0_A+3p^0_B$ ...(i)
or
Total vapour pressure of solution (1 mole A + 4 mole B)

$=\dfrac{1}{5}p^0_A+\dfrac{4}{5}p^0_B$

$560=\dfrac{1}{5}p^0_A+\dfrac{4}{5}p^0_B$

$2800=p^0_A+4p^0_B$ ...(ii)
or
Solving eqs. (i) and (ii),

$p_B^0=600$ mm of Hg = vapour pressure of pure B

$p_B^0=400$ mm of Hg = vapour pressure of pure A

Match the items of Column-I with its proportional term in the items of Column-II:
Column-I Column-II
(a) Kinetic energy (p) Mole fraction
(b) Partial pressure of a gas (q) Density
(c) Rate of diffusion (r) Molar mass
(d) Vapour pressure of a liquid (s) Absolute temperature

a) According to kinetic molecular theory of gasesKinetic energy is directly proportional to absolute temparature.b) A ccording to Henry's law Partial pressure of a gas is directly proportional to mole fraction.
c) Rate of diffusion is inversely proportional to density.
d) vapour pressure of liquid is directly proportional to molar mass of liquid.

The answer to this question is a single digit integer, ranging from 0 to 9.

The van't Hoff factor for aqueous solution of $CuSO_4 \cdot 5H_2O$ will be ...

The van't Hoff factor, i, is the number of particles formed in a solution from one formula unit of solute.

$CuSO_4\rightarrow Cu^{+2}+SO_4^{-2}$
So the total number of ions are 1+1 that is 2
So vantoff factor for the above compound is 2

A solution of a non-volatile solute in water freezes at $-0.3^0 C$ .If vapour pressure of pure water at 298 K is 23.51 mm of Hg and $K_f$ for water in 1.86 degree/molal .Calculate the vapour pressure of this solution at 298 k.

1309016_711706_ans_31ace6f19f534d0eb453bf7b4e226d30.jpg

At 80C, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole per cent which at 80C is in equilibrium with the vapour containing 30 mole per cent of benzene.

$P_B^o =$ vapour pressure of pure Benzene

$=753\ mm\ Hg$

$P_1^o =$ vapour pressure of pure Tolvene

$=290\ mm\ Hg$

$x_B=$ mole frection of benzene above the liquid

$=0.30$

Now, potential pressure of benzene above the liquid

$P_B =x_B \times P_{Total}$

$\Rightarrow \ 0.30=\dfrac {P_B}{P_{Total}}.....(1)$

Now, Also, potential pressure of benzene above the liquid using Roonth is law

$P_B =x_{B\ liquid} \times P_B^o$

Similarly potential pressure of Tohve above the liquid

using Raonlt's law

$P_T=X_{T\ liquid} \times P_{T}^o= (1-x_{B\ liquid})P_T^o$

$\therefore \ P_{Total}=x_{B\ liquid}\times P_B^o +(1-x_{B\ liquid})P_T^o$

$=P_T^o +(P_B^o -P_T^o)x_{B\ liquid}$

using

$(1)$

$0.30=\dfrac {P_B}{P_T^O (P_B^o -P_T^o)x_{B\ liquid}}=\dfrac {x_{B\ liquid} P_B^o}{P_T^o +(P_B^o -P_T^o)x_{B\ liquid}}$

Solving, we get

$x_{B\ liquid}=\dfrac {0.30\ P_T^o}{0.70\ P_B^o -0.30\ P_T^o}=\dfrac {0.30\times 290}{(0.70\times 753 -0.30\times 290)}$

$=0.142$

$x_{T\ liquid}=0.858$

$\therefore \$ Mole percent benzene

$=14\%$

Mole percent Tohene

$=86\%$

Two solids $X$ and $Y$ dissociate into gaseous products at a certain temperature as follows:
$X(s) \rightleftharpoons A(g) + C(g)$ , and $Y(s) \rightleftharpoons B(g) + C(g)$ . At a given temperature, pressure over excess solid $X$ is $40\ mm$ and total pressure over solid $Y$ is $60\ mm$ . When they are preset in separate containers. Calculate
(a) the values of $K_{p}$ for two reactions (in mm)
(b) the ratio of moles of $A$ and $B$ in the vapour state over a mixture of $X$ and $Y$ .
(c)The total pressure of gases over a mixture of $X$ and $Y$ .

1309018_785217_ans_74ed8662e4754f59b623b88bc827d750.jpg

Two solutions of A and B are available. The first is known to contain 1 mole of A and 3 moles of B and its total vapour pressure isAtm. The second is known to contain 2 moles of A and 2 moles of B; its vapour pressure is greater than 1 atm, but it is found that this total vapour pressure may be reduced to 1 atm by the addition of 6 moles of C. the vapour pressure of pure C is 0.80 atm. Assuming ideal solutions and that all these data refer to 25C, calculate the vapour pressureof pure A and of pure B.

solution

$1^{st}$

$n_A =1\ mole; \ n_B=3\ mole$

$\therefore \ x_A=\dfrac {1}{1+3}=0.25$

$x_B=\dfrac {3}{1+3}=0.75$

$\therefore \$ Raolt's Law

$P=P_A^o x_A+P_B^o x_B$

$\Rightarrow \ 1=0.25\ P_A^o +0.75\ P_B^o ....(1)$

solution

$2^{nd}$ ;

$x_A=\dfrac {2}{2+2+6}=0.2;\ x_B=\dfrac {2}{2+2+6}=0.2;\ x_c=\dfrac {6}{2+2+6}=0.6$

$\therefore \$ Raolt's Law

$1=0.20\ P_A^o +0.20\ P_B^o +0.60\ P_C^o .......(2)$

$\Rightarrow \ 1=0.20\ P_A^o + 0.20\ P_B^o +0.60 \times 0.8$

$\Rightarrow \ 0.52=0.20 (P_A^o +P_B^o)$

$\Rightarrow \ P_A^o +P_B^o =2.6......(3)$

From

$(1)$ &

$(3)$

we get

$P_A^o =1.9\ atm$

$P_B^o =0.7\ atm$

Vapour pressure of water at certain temperature is $155$ mm Hg and that of the another solvent 'X' is 'p' mm Hg. Molecular weight of 'X' is $128$ . An aqueous solution of 'X'( $64\%$ by wt) has a vapour pressure of $145$ mm Hg. What is 'p'?

vapour pressure of water =

$P_A^o = 155\ mm\ Hg$

vapour pressure of solvent

$P_B = p$

Let

$100$ g aqueous solution

Amount of solvent =

$64 g$

Amount of water =

$36 g$

number of mole of water

$n_A = \dfrac{36}{18} = 2$

number of mole of solvent

$n_B = \dfrac{64}{128} = 0$

mole fraction

$X_A = \dfrac{2}{2 + 0.5} = 0.8$

$X_B = 0.2$

$P_A^0 x_A + P_B^0 x_B = 145$

$155 \times 0.8 + P \times 0.2 = 145$

$0.2 P = 145 - 124$

$P = \dfrac{21}{0.2} = 105 \, mm\ Hg$

Vapour pressure of chloroform $(CHCl_{3})$ and dichloromethane $(CH_{2}Cl_{2})$ at $298\ K$ are $200\ mm\ Hg$ and $415\ mm\ Hg$ respectively.
(i) Calculate the vapour pressure of the solution prepared by mixing $25.5\ g$ of $CHCl_{3}$ and $40\ g$ of $CH_{2}Cl_{2}$ at $290\ K$ and
(ii) Mole fraction of each component in solution.

Vapour pressure of pure chloroform =

$P_A^o = 200 mm$

Vapour pressure of pure dichloromethane

$P_B^o = 415 \, mm Hg$

Molar mass

$M_A = 119.5 g$

Molar mass

$M_B = 85 \, g$

(i) mass of

$CHCl_3 = 25.5 g$

number of mole

$= \dfrac{25.5}{119.5} = 0.213$

Mass of

$CH_2Cl_2 = 40 g$

Number of mole

$= \dfrac{40}{85} = 0.47$

$x_A = \dfrac{0.213}{0.213 + 0.47} = 0.322$

$x_B = 1 - 0.322 = 0.688$

$P = P_A^o X_A + P_B^o x_B$

$= 200 \times 0.322 + 415 \times 0.688$

$= 64.4 + 285.52 = 349.92$

(ii) Mole fraction of chlotoform =

$0.322$

Mole fraction of

$CH_2 Cl_2 = 0.688$

Calculate the weight of a non-volatile solute $(mol./wt. 40)$ which should be dissolved in $114\ g$ octane to reduce its vapour pressure to $80$ %.

Weight of non volatile solute m

mol

$wt . M = 40$

octane

$(C_8 H_{18} = 8 \times 12 + 18 = 114 g)$

No. of mole

$=$

$\dfrac{114}{114} = 1$ mole

Number of mole of octane

$= 1$ mole

Number of mole of solute

$=$

$\dfrac{m}{40}$

mole fraction

$=$

$\dfrac{1}{1 + \dfrac{m}{40}} = \dfrac{40}{40 + m}$

$P_A^o \dfrac{40}{40 + m} = \dfrac{P_A^o \times 80}{100}$

$50^o = 40 + m$

$m + 10\ g$

Vapour pressure of pure benzene is $640\ mm$ and its mol. wt. is $78$ . When $2.175\ g$ of a non-volatile substance is dissolved in $39g$ benzene the vapour pressure of the solution becomes $600\ mm$ . Calculate mol. wt. of solute.

vapour pressure of pure Benzene =

$P_A^{\circ} = 640 \, mm$

Mol. wt of Benzene =

$78^o$

mole of non volatile substance = n

mole of Benzene =

$\dfrac{39}{78} = 0.5$

mole fraction =

$\dfrac{n}{n + 0.5}$

$P_A^o - P_A^o . \dfrac{n}{n + 0.5} = 600$

$640 - 600 = 640 \dfrac{n}{n + 0.5}$

$\dfrac{40}{640} = \dfrac{n}{n + 0.5}$

$n + 0.5 = 16 n$

$15.5 n = 0.5$

$n = \dfrac{0.5}{15.5}$

$\dfrac{2.175}{M} = \dfrac{0.5}{15.5}$

$M = \dfrac{2.175 \times 15.5}{0.5}$

$= 67.425$

Two gases in adjoining vessels are brought into contact by opening a stop cock between them. The one vessel measured 0.25 lit and contained NO at 800 torr and 220 K, the other measured 0.1 lit and contained $O_2$ at 600 torr. The reaction to form $N_2O_4$ (solid) exhausts the limiting reactant completely.
a) Neglecting the vapour pressure of $N_2O_4$ what is the pressure of the gas remaining at 220 K after completion of the reaction?
b) What weight of $N_2O_4$ is formed?
[Given that : $\displaystyle \frac {342.8}{760}\,\times\, \frac {0.35}{0.082}\,\times\, \frac{1}{22}\,=\,8.75\,\times\,10$ ]

number of mole of No in 1st vessel

$n_1 = \dfrac{PV}{RT}$

$= \dfrac{(800 \, torr)(0.25 \, lit)}{R \times (220\ k)}$

number of mole

$O_2$ in

$II^{nd}$ vessel

$n_2 = \dfrac{PV}{RT}$

$= \dfrac{(600 \, torr)(0.1 \, lit)}{R \times (220\ k)}$

$2NO + O_2 \rightarrow N_2O_4$

After completion of this reaction

$O_2$ is limiting reapet (by comparing

$n_1$ and

$n_2$ )

number of mole of NO left

$= n_1 - 2n_2$

Number of mole of

$N_2 O_4$ formed =

$n_2$

$(1)$ pressure

$= \dfrac{(n_1 - 2n_2)V}{R \times 220\ k}$

$= \dfrac{-\dfrac{ 1}{R \times 220 k} (800 \times 0.25 - 2 \times 600 \times 0.1)}{R \times 220 k} \times 0.35$

$= 80 \, torr \times 0.35$

$= 28 \, torr$

$(2)\ n_2 = \dfrac{600 \times \dfrac{1}{760} \times 0.1}{0.0821 \times 220} = 0.00437$

Mass =

$0.00437 \times 92 = 0.40 g$

The vapour pressure of chloroforms and de-chloroforms at 298 k are 200 mm Hg and 415 mm Hg? calculate the vapour pressure of solution by mixing 25.5 gram of $CHCl_3$ 40 gram of $CH_2Cl_2$ . calculate the mole fraction of each component on vapour phase.

$P_A^0$ (choloroforms)

$=200\ mm\ Hg$

$n_A=\dfrac{25.5}{(12+1+35.5\times 3)}=0.2134$

$P_0^B(CH_2Cl_2)=415\ mm\ Hg$

$n_B=\dfrac{40}{(12+2+31.5\times 2 )}=0.4706$

$X_A=\dfrac{0.2134}{0.2134+0.4106=}\dfrac{0.2134}{0.684}$

$=0.312$

$X_B=0.688$

$P_A=P_A^oX_A=200\times 0.312=62.4$

$P_B=P_B^o X_B=0.685\times 415=285.52$

$P=62.4+285.52=347.92$

Mole fraction of

$CH_2Cl_2$ in vapour phase

$=\dfrac{285.52}{347.92}-0.8206$

$\simeq 0.82$

Mole fraction of

$CHCl_3$ in vapour phase

$=\dfrac{62.4}{347.92}=0.18$

Van't Hoffs factor for a solution is less than one, what is the conclusion drawn from it.

Van't Hoff factor for a solution is less than one, the conclusion drawn from it is that the solution undergoes association.

The Van't Hoff factor

$\displaystyle i = \dfrac { \text { number of solute particles present in solution } }{\text { theoretical number of solute particles due to dissolution of non electrolyte }}$

$\displaystyle i = \dfrac { \text { n observed} }{\text {n theoretical }}$

Van't Hoff factor for a solution is less than one

$\displaystyle i = \dfrac { \text { n observed} }{\text {n theoretical }} < 1$

$\displaystyle = \text { n observed} < \text {n theoretical }$

This suggests association.

Note: For dissociation,

$\displaystyle i > 1$ .

At $400^oC$ , the vapour pressure of water is $55.3mm\ Hg$ . Calculate the vapour pressure at the same temperature over $10\%$ aqueous solution of urea $[CO(NH_2)_2]$ ?

$\dfrac{p^0 - p}{p^0} = X_{urea}$

let the mass of solution be 100g

mass of urea =10g

M.wt of urea = 60; M.wt of water = 18

$X_{urea} = \dfrac{10}{60} = \dfrac{1}{6}$

$X_{water} =\dfrac{90}{18} =5; X_{urea} =\dfrac{1/6}{1/6+5} = \dfrac{1}{31}$

mole fraction of urea:

$= \dfrac{55.3 -p}{55.3} = \dfrac{1}{31}$

$p = 55.3* \dfrac{30}{31} = 53.52 mmHg$

The vapor pressure of water at $80^ {\circ}C$ is $355$ torr. A $100\, ml$ vessel contained water-saturated oxygen at $80^ {\circ}C$ the total gas pressure being $760$ torr. The contents of the vessel were pumped into a $50.0$ ml, vessel at the same temperature, What were the partial pressure of oxygen and of water vapor, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense.

Total pressure

$= {p_{{O_2}}}^{\rm{o}} + {p_{{H_2}O}}^{\rm{o}}$

$\begin{array}{l}760 = {p_{{O_2}}}^{\rm{o}} + 355\\{p_{{O_2}}}^{\rm{o}} = 405\,{\rm{torr}}\end{array}$
Since,

${p_1}{v_1} = {p_2}{v_2}$

$\begin{array}{l}{p_{{H_2}O}}^{\rm{o}} = \dfrac{{{p_{{O_2}}}^{\rm{o}} \times 100}}{{50}} = \dfrac{{405 \times 100}}{{50}} = 810\\{p_{{H_2}O}}^{\rm{o}} = 810\,{\rm{torr}}\end{array}$
total pressure in

$50$ ml contain

$=810+355$

$=166\, {\rm{torr}}$

Three grams of activated charcoal was added to 50ml of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. Amount of acetic acid adsorbed (per gram of charcoal) is:

Initial millimoles of acetic acid=

$NV$

$= 50 \times 0.06$

$=3 mmol$

Final millimoles of

$CH_3COOH= 50 \times 0.42$

$= 2.1mmol$

Acetic acid adsorbed=

$3-2.1=0.9 mmol$

Weight after

$60 min=\cfrac {0.9}{10^3}\times 60= 54mg$

$\therefore$ Amount adsorbed per

$mg$ of charcoal=

$\cfrac {54}{3}=18 mg$

Vapour pressure of pure water at $298\ K$ is $23.8\ mm\ Hg$ . $50\ g$ of urea $(NH_2CONH_2)$ is dissolved in $850\ g$ of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapour pressure of water,

$P_1=23.8$ m of hg

The weight of water=850 g

The weight of Urea=50 g

The molecular weight of water(H_2O)=

$1\times 2+16=18\ g\ mol^{-1}$

molecular weight of urea

$(NH_2CONH_2)=2N+4H+C+O=2\times 14+4\times 1+12+16=60\ g\ mol^{-1}$

Use formula

number of moles=

$\cfrac{\text{mass}}{\text{molar mass}}$

number of moles of water

$n_1$ =

$\cfrac{850}{18}$ =

$47.22$

number of moles of urea

$n_2$ =

$\cfrac{50}{60}$ =0.83

Now, we have to calculate the vapour pressure of water in the solution. we take vapour pressure as

$P_1$ .

Use the formula of Raoult's law

$\cfrac{P_1^0-P_1}{P_1^0}=\cfrac{n_2}{n_1+n_2}$

plug the values we get

$\cfrac{23.8-P_1}{23.8}=\cfrac{0.83}{47.22+0.83}$

$\cfrac{23.8-P_1}{23.8}=0.0173$

$23.8-P_1$ =

$23.8\times 0.0173$

$P_1=23.4$ m Hg

Vapour Pressure of water in the given solution

$=23.4$ mm of Hg

Relative lowering=

$0.0173$

Two liquids $A$ and $B$ form an ideal solution at temperature $T$ . When the total vapour pressure above the solution if $600\ torr$ , the amount fraction of $A$ in the vapour phase is $0.40$ and in the liquid phase is $0.6$ . What are the vapour pressure of pure $A$ and $B$ at temperature $T$ ?

Let

$P_{A}^{\circ}$ be vapour pressure of pure

$A$ at temperature

$T$ .

Fraction of A in vapour phase

$=\cfrac{P_{A}}{P_{total}}$

Let

$x_{A}$ be mole fraction of A in liquid phase.

$\therefore 0.4=\cfrac{x_{A}\times P_{A}^{\circ}}{P_{total}}=\cfrac{0.6\times P_{A}^{\circ}}{600}$

$\Rightarrow P_{A}^{\circ}=400\,Torr$

Also,

$P_{total}=P_{A}^{\circ}\cdot x_{A}+P_{B}^{\circ}\cdot x_{B}$

$\therefore 600=400\times 0.6+P_{B}^{\circ}\times (1-0.6)$

$\therefore P_{B}^{\circ}=\cfrac{600-240}{0.4}=900\,Torr$

Hence,

$P_{A}^{\circ}=400\,Torr, P_{B}^{\circ}=900\,Torr$

Solution of two volatile liquid $x$ and $y$ obey Raoult's law. At a certain temperature it is found that when the total pressure above a given solution is $400\ mm$ of $Hg$ , the mole fraction of $x$ in the vapour is $0.45$ and in the liquid is $0.65$ . what are the vapour pressure of two pure liquids at the given temperature?

According to Dalton's Law of Partial Pressures, Partial Pressure of each component gas is proportional to their mole fractions in the vapour.

Hence,

Partial Pressure of A

$= Pa = 400\times 0.45 = 180$

$Pb = 400\times(1-0.45) = 220$
Also, Partial Pressure of the gas in the vapour = (Vapour Pressure) × (Mole Fraction of Liquid in solution)

Hence,

Vapour Pressure of A

$= Va = \dfrac{180}{0.65} \approx 277\ mm\ of\ Hg$

$Vb = \dfrac{220}{1-0.65} \approx 629\ mm\ of\ Hg$

The vapour pressure of two pure liquids $A$ and $B$ which from an ideal solution are $500$ and $800$ torr respectively at $300\ K$ . A liquid solution $A$ and $B$ for which the mole fraction of $A$ is $0.60$ is contained in a cylinder closed by a piston on which the pressure can be varied. What will be the pressure when 1 mol of the mixture has been vaporized?

Let

${ n }_{ B }$ moles of B is present in 1 mole of mixture that has been vapourized.

Thus,

${ y }_{ B }=\dfrac { { n }_{ B } }{ 1 }$

Mole fraction of B in the remaining liquid phase is given by:

${ \partial }_{ B }=\dfrac { 1-{ n }_{ B } }{ 1 }$

${ \chi }_{ B }=\frac { p-{ p }^{ 0 }r }{ { p }_{ B }^{ 0 }-{ p }_{ r }^{ 0 } }$

$[\because { p }^{ 0 }r+({ p }_{ B }^{ 0 }-{ p }_{ r }){ \chi }_{ B }]$

${ y }_{ B }=\dfrac { { p }_{ B } }{ p } \Rightarrow { p }_{ B }^{ 0 }-{ p }_{ r }^{ 0 }$

After substituting of values of

${ \chi }_{ B }$ :

$1-{ n }_{ B }=\dfrac { p-{ p }^{ 0 }r }{ { p }_{ B }^{ 0 }-{ p }_{ r }^{ 0 } }$

and

${ n }_{ B }$ =

$\dfrac { (1-{ n }_{ B }){ p }^{ 0 }B }{ p }$

${ n }_{ B }=\dfrac { { p }^{ 0 }B }{ p+{ p }^{ 0 }B }$

So, 1-

$\dfrac { { p }^{ 0 }B }{ p+{ p }^{ 0 }B }$ =

$\dfrac { p-{ p }^{ 0 }r }{ { p }_{ B }^{ 0 }-{ p }_{ r }^{ 0 } }$

$\Rightarrow p=\sqrt { { p }_{ B }^{ o }.{ p }_{ r }^{ o }=\sqrt { 100X900 } }$

$\Rightarrow p=300\ torr$ is the answer.

Vapor pressure of pure water at $298 \ K$ is $23.8 \ mmHg$ . $50 \ g$ of urea is dissolved in $850 \ g$ of water. Calculate the vapor pressure of water for this solution and it's relative lowering.

$P^o_A=23.8 \ Mm \ Hg; \ m_{H_2O}=850 \ gm; \ m_{urea}=50 \ gm$

Moles of $H_2O=47.22 \ moles$

Moles of Urea $= \cfrac {50}{60}=0.833$

Mole fraction of $H_2O= \cfrac {47.22}{48.053}=0.982$

Vapor pressure of water $= (mole \ fraction) \times P^o_A$ (Raoults Law)

$P_A=0.982 \times 23.8 = 23.37 \ mm \ Hg$

Relative lowering $= \cfrac {P^o_A P_A}{P^o_A}=0.018 \ mm \ Hg$

Two liquids $A$ and $B$ boils at $145^{o}C$ and $190^{o}C$ . Which of them has higher vapour pressure at $80^{o}C$ ?

Vapour pressure correlates with boiling point inversely. In this case liquid A has a lower boiling point and thus will have higher vapour pressure. Since liquid A has a lower boiling point you can assume that its inter-molecular forces are weaker. Therefore, it is much more likely to become a vapour at lower temperatures and more vapour means more pressure.

The vapour pressure of pure water is $23.8\ mmHg$ at $25^{0}\ C$ . What is the vapour pressure of $2.50\ molal\ C_{6}H_{12}O_{6}$ ?

vapour pressure of water

$P_A^O=23.8\ mm\ Hg$

molarity of solution

$=2.5\ molal$

Let amount of water

$=1\ kg$

number of mole of

$C_6H_{12}O_6=2.5$

number of mole of water

$=\dfrac {1000}{18}$

mole fraction of water

$X_A=\dfrac {\dfrac {1000}{18}}{\dfrac {1000}{18}+2.5}=\dfrac {1000}{1045}$

Vapour pressure

$P_A=P_A^O X_A$

$=23.8\times \dfrac {1000}{1045}=22.775\ mm\ Hg$

The normal boiling of water is $373 K$ . Vapour pressure of water at temperature $T$ is 19 mm $Hg$ . If enthalpy of evaporation is $40.67 kJ/mol$ , then temperature $T$ would be:
(Use $log 2 =0.3,R =8.3 { JK }^{ -1 }$ )

By Clausius- Clayperon Equation

$ln\cfrac {P_2}{P_1}=\cfrac {\Delta H}{R}\left[\cfrac {1}{T_1}-\cfrac {1}{T_2}\right]$

Here,

$T_2= 373K, P_2= 760 mm Hg, T_1=T$ &

$P_1=19$

$\therefore ln\left(\cfrac {760}{19}\right)=\cfrac {40670}{8.3}\left[\cfrac {1}{T}-\cfrac {1}{373}\right]$

$\Rightarrow T= 291.3K$

For a dilute solution containing 2.5 grams of non volatile non electrolyte solute in hundred grams of water, the elevation in the boiling point at 1 atmosphere pressure is to degree Celsius asuming concentration of solute is much lower than concentration of solvent, the vapour pressure of the solution is?

According to

Elevation of Boiling Point $\Delta {{T}_{b}}=i\ {{K}_{b}}m$

Where,

i= Vant’s Hoff factor,

Kb= Boiling point Elevation Constant

m= molarity

So,

$Molarity=\dfrac{\Delta {{T}_{b}}}{{{K}_{b}}}$

$\ \ \ \ \ \ \ \ \ \ \ \ =\dfrac{2}{0.76}=2.63\ mol\ K{{g}^{-1}}$

Now,

We know that the molarity of pure water $=55.6\ mol\ K{{g}^{-1}}$

Also given that

Atmospheric pressure of $760\ mm\ Hg\ at\ {{100}^{0}}C$

So,

$\dfrac{{{P}_{0}}-P}{{{P}_{0}}}=\dfrac{n}{N}$

$\dfrac{760-P}{760}=\dfrac{2.63}{55.56}$

$P=724\ torr$

Hence,

The vapour pressure of solution $P=724\ torr$

This is the required solution.

Quantity of gas occupies $100m$ ,then it is collected on surface of water at $15C$ and $760mm$ of $Hg$ . It occupies $91.9ml$ in dry state at STP conditions. Calculate vapour pressure of ${H}_{2}O$ at $15C$ .

1353205_1153619_ans_2c36fa2149ef422e80cee76b4d43aa3f.jpg

By adding a non-volatile salute into a solution vapour pressure is reduces.

lowering in vapour pressure occur by adding a non volatile solute into a solution

$P^o$

$\rightarrow$ vapour pressure of pure liquid

$P$

$\rightarrow$ vapour pressure of solution

$P^o-P$

$\rightarrow$ lowering in vapour pressure

$\cfrac {P^o-P}{P^o}$

$\alpha$

$X_b$

$\rightarrow$ is mole fraction of solute.

Give the differences between colloidal solution, true solution and suspension? (at least 2 differences)

1693303_1277601_ans_0ad7788f272f4996a7f8c91811945105.PNG

Liquids $'A'$ and $'B'$ form an ideal solution. Calculate the vapour pressure of solution having 40 mole-percent of $'A'$ in the vapour at equilibrium.
$(P^0_A = 80cm Hg, P^0_B = 30cm Hg)$

$\dfrac { 1 }{ P_{ total } } =\dfrac { Y_{ A } }{ P_{ A }^{ 0 } } +\dfrac { Y_{ B } }{ P_{ B }^{ 0 } } =\dfrac { 0.4 }{ 80 } +\dfrac { 0.6 }{ 30 } =\dfrac { 1 }{ 40 }$

$P_{total} = 40 \ cm\ Hg$

Solve it

1356144_1283272_ans_d64a71c6f12541db95c0478271e6b6c2.jpg

Urea forms an ideal solution in water. Determine the V.P of an aqueous solution containing $10\%$ by mass of urea at $40^\circ C$ . (Vapour pressure of water at $40^oC=55.3 mm Hg)$

2040212_1289490_ans_70f0a166995747f98171ea67025e0feb.jpg

${P_T} = P\mathop A\limits^ \circ {X_P} - P\mathop B\limits^ \circ {X_B}$
$P_T = P \mathring{A} X_P - P \mathring{B} (1 - X_A)$
${P_T} = P\mathop A\limits^ \circ {X_A} + P\mathop B\limits^ \circ - P\mathop B\limits^ \circ {X_A}$
${P_T} = P\mathop B\limits^ \circ + \left( {P\mathop {A - P\mathop B\limits^ \circ }\limits^ \circ } \right){X_A}$

1352481_1281773_ans_58e7febf8371429eb36f9bfc03951c68.jpg

0.6 ml of acetic acid $(C{H_3}COOH)$ having density 1.06 g/ml is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was $0.0205\,{\,^0}C.$ Calculate the Van't Hoff factor and the dissociation constant of acid.

2041204_1358362_ans_b4e778d5807b41c2ab4a93aa586af874.jpg

Why is the vapour pressure of a solution of glucose in water lower than that of water?

1657295_1555320_ans_270514c44e344565854376ad783c3a1c.jpg

A quantity of 1.245 g of $CuSO_{4}.xH_{2}O$ was dissolved in water and $H_{2}S$ was passed into it till CuS was completely precipitated. The $H_{2}SO_{4}$ produced in the filtrate required 10 ml of M - NaOH solution. Calculate x.

We have Reaction

$CaSO_{4}.xH_{2}O+H_{2}S \rightarrow CuS + H_{2}SO_{4}+XH_{2}O$ .
1 mass

$\rightarrow$ 1 mole

mole

$\Rightarrow \dfrac{mass}{molar\ mass}\rightarrow \dfrac{1.245}{\left ( 159.6+18x \right )}$

$\left [ ml. at . of\ CaSO_{4}.xH_{2}O=\left ( 159.6+18 x \right ) \right ]$

Also, we have neutralisation reaction

$H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O$ .

1 mole

$\leftarrow$ 2 mole.

$= \dfrac{0.01}{2} \leftarrow 0.01\times 1$

$=$ 0.005 mole

$H_{2}SO_{4}$

$\Rightarrow \dfrac{1.245}{159.6+18x} = 0.005$

On solving we get

$x = 4.967$

$x \approx 5$ - single integer

Answer x = 5

A sample of chalk contain as impurity a form of clay which losses $14.5\%$ if its weight as water on along heating. A $5\ g$ of chalk sample, on heating shows a loss in weight by $1.507\ g$ . The mass percentage of $CaCO_3$ in the chalk sample is $(Ca=40)$

In the question given

$5\ gm$ of chalk sample
mass of moisture lost By clay

$=()5-x\times \dfrac {14.5}{100}$

$72.5-14.5\ y$

$-725-145x$
By Heating

$CaCO_3$ lost

$CO_2$ also

$CaCO_3\xrightarrow [ ]{ \Delta } CaO +CO_2$

$100\quad \quad \quad 44\ gm$
so

$100\ gm\ CaCO_3 \$ lost

$44\ gm\ CO_2$

$44\ gm\ CO_2$ lost By

$100\ gm\ CaCO_3$
So

$x\ gm$ of

$CO_2$ lost By

$=44\ x$ gram
total mass of lost

$=1.507\ gm$

$725-145x+44x=1.507$

$295x=782$

$x=2.65$
so, gram of

$CO_2=44\times 2.65$
moles of

$CO_2=1.16/44=0.02$
mass of

$CaCO_3=0.2\times 100=2.63$

$\%$ of

$CaCO_3$ in sample

$\dfrac {2.63}{5}\times 100=53/g$

A quantity of $2.7\ g$ of an alloy of copper and silver was dissolved in moderately concentrated $HNO_{3}$ , and excess of $HCl$ was added to this solution when $2.87\ g$ of a dry precipitate is formed. Calculate the percentage of copper in the alloy.
$(Cu=63.5\ ,\ Ag=108)$

1975403_1682620_ans_8384f53feb164941be4078d3b3cda6fc.jpg

A gas mixture contains $CH_4$ and $C_3H_6$ . When this mixture undergo cracking into $C(s)$ and $H_ 2(g)$ m the total number of moles of $H_2(g)$ obtained is $42$ . If the total volume of the initial gas mixture at $1.5\ atm$ and $27^oC$ is $246.3\ L$ , what is the mole per cent of $CH_4$ gas in the initial mixture?

$CH_{4}\Rightarrow C+2H_{2}$ Let

$n_{H_{2}}$ from

$CH_{4}$ be

$x$

$1$ mole

$2$ mole

$C_{3}H_{6}\rightarrow 3C+3H_{2}$

$\therefore\ n_{H_{2}}$ from

$C_{3}H_{6}$ be

$42-x$

$\therefore \ ^{n}CH_{4}=\dfrac{H}{2}; \ ^{n}C_{3}H_{6}=\dfrac{42-x}{3}$
Now,

$PV=nRT$

$n=\dfrac{PV}{RT}=\dfrac{1.5\times 246.3}{0.0821\times 300}=15\ mole (CH_{4}+C_{3}H_{6})$

$\therefore \dfrac{H}{2}+\dfrac{42-x}{3}=15\Rightarrow 3x+2(42-x)=90$

$n=90-84=6\ mole$

$\therefore \ ^{n}CH_{4}=\dfrac{H}{2}=3\ mole$

$\therefore$ mole

$\%$ of

$CH_{4}=\dfrac{3}{15}\times 100=20\%$
Ans will be

$0020$

To analyse cast iron for its sulphur content, a $6.4\ g$ portion of the iron was weighed out for analysis and treated as follows: it was dissolved in hydrochloric acid, the hydrogen sulphide evolved from iron sulphide was distilled off and made to be absorbed by a solution of a cadmium salt, after which $CdS$ was treated with an excess of a solution of $CuSO_{4}$ and the $CuS$ precipitated formed was ignited. As a result $0.795\ g$ of an ignited $CuO$ precipitate was obtained. Calculate the percentage content of sulphur in the cast iron $(Cu=63.5)$

$FeS+2HCl \longrightarrow FeCl_2+H_2S\uparrow$

$FeS+Cd^{2+}(aq) \longrightarrow CdS+Fe^{2+(aq)}$

$Cds+CuSO_4 \longrightarrow CuS + CdSO_4$

$2CuS + 3O_2 \longrightarrow 2CuO + 2SO_2$

Mole of

$CuO=\dfrac{0.795}{79.5}=0.01\ moles$

Mass of

$S=0.01 \times 32 = 0.32g$

$\%$ content of

$S$ in

$FeS=\dfrac{0.32g}{6.4g}\times 100$

$=5$

A mixture containing $As_{2}S_{3}$ and $As_{2}S_{5}$ requires 20 ml of 0.05 N iodine for titration. The resulting solution is then acidified and excess of KI was added. The liberated iodine required 1.24g hypo, $Na_{2}S_{2}O_{3}.5H_{2}O ,$ for complete reaction.The reactions are :
$As_{2}S_{3}+2I_{2}+2H_{2}S \rightarrow As_{2}S_{5}+4H^{+}+4I^{-}$
$As_{2}S_{5}+4H^{+}+2I^{-}\rightarrow As_{2}S_{3}+2I_{2}+2H_{2}S$
The mole percent of $As_{2}S_{3}$ in the original mixture is (As= 75)

1976511_1682709_ans_98108e17f3c7455b9a84f0144867d349.jpg

A 0.2 g sample of chromite was fused with excess of $Na_{2}O_{2}$ and brought into solution according to reaction:

$2Fe\left ( CrO_{2} \right )_{2}+7Na_{2}O_{2}$ $\rightarrow 2NaFeO_{2}+4Na_{2}CrO_{4}+2Na_2O$

The solution was acidified with dil. HCl and 1.96 g Mohr's Salt (molar mass = 392 g/mol ) was added. The excess of $Fe^{2+}$ required 40 ml of $0.05N - K_{2}Cr_{2}O_{7}$ for titration. What is the percent of Cr in sample ? (Cr = 52, Fe = 56)

2036657_1682694_ans_9ca8242eab42402cb64896b1d297a2e6.jpg

A 10g mixture of $Cu_{2}S$ and CuS was treated with 400 ml of $0.4M - MnO_{4}^{-}$ in acid solution producing $SO_{2},Cu^{2+}$ and $Mn ^{2+}.$ The $SO_{2}$ was boiled off ans the excess of $MnO_{4}$ was titrated with 200ml of $1M-Fe^{2+}$ solution. The percentage of CuS in original mixture is (Cu=64)

2036658_1682697_ans_ee123dd52be04ff195bc8cf21b815743.jpg

A quantity of $1.6\ g$ of pyrolusite ore was treated with $50\ ml$ of $1.0\ N$ oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to $250\ ml$ in a flask. A volume of $25\ ml$ of this solution when titrated with $0.1\ N\ KMnO_4$ required $32\ ml$ of the solution. The percentage of available oxygen in the ore is:

1975405_1682657_ans_c5b0288132f34711a661db7bf63f66fd.jpg

One gram of commercial $AgNO_{3}$ is dissolved in 50 ml of water. It is treated with 50 ml of a KI solution.The silver iodide thus precipitated is filtered off. Excess of KI is titrated with $M/10-KIO_{3}$ solution in the presence of 6M - HCl till all iodide ions are converted into ICl. It requires 50 ml of $M/10-KIO_{3}$ solution. A 20 ml of the same stock solution of KI requires 30 ml of $M/10-KIO_{3}$ under similar conditions. The percentage of $AgNO_{3}$ in the sample is (Ag=108)
Reaction:
$KIO_{3}+2KI+6HCl$
$\rightarrow 3ICI+KCI+3H_{2}O$

1976566_1683010_ans_6c52f9d1d99e40b6aecfc9dadc80b09c.jpg

A mixture of ideal gases is cooled up to liquid helium temperature (4.22 K) to from an ideal solution. Is this statement true or false? Answer '1' or true and '2' for false.

1978827_1683975_ans_3f39cea1730f4b6ba939c17a1d354dbe.jpg

$50\ mL$ if a mixture of $CO$ and $H_{2}$ gave $20\ mL$ of $CO_{2}$ after combustion in excess of air. Determine the percentage of $CO$ by volume in the mixture.

$CO + H_2 + O_2$

$\xrightarrow{Combustion}{CO_2 + H_2O}$

This implies that

$n$ moles of

$CO$ shall give

$n$ moles of

$CO_2$ .

Since we get

$20\;mL$ of

$CO_2$ post combustion, hence

$CO$ initially must be

$20\;mL$ .

Thus, percentage of

$CO$ by volume in the mixture =

$\frac{20}{50}$ X 100 =

$40$ %

$[Answer]$

The vapour pressure of an aqueous solution of cane sugar $(mol. wt\ 342 )$ is $756\ mm$ at $100^{o}C$ . How many grams of sugar are present per $1000\ g$ of water?
Give your answer in nearest integer.

$X_{Solute}=\dfrac{760-756}{760}=\dfrac{4}{760}$

$\Rightarrow \dfrac{n_{solute}}{n_{solvent}}=\dfrac{4}{756}\left(\because \dfrac{n_{S}}{n_{solvent}}=\dfrac{P^{0}_{solvent}-P_{Solution}}{P_{solution}}\right)$

$\Rightarrow \dfrac{\dfrac{wt\ solute}{342}}{\dfrac{1000}{18}}=\dfrac{4}{756}$

Wt of solute

$=\dfrac{4\times 342\times 1000}{756\times 18}=100.53\ g$

Ans-

$100$

What approximate proportions by volumes of water $(d=1\ g/cc)$ and ethylene glycol $C_{2}H_{6}O_{2}(d=1.12\ g/cc)$ must be mixed ot ensure protection of an automobile cooling system to $-10^{o}C$ ?

Let

$\dfrac{V_{water}}{V_{C_{2}H_{6}O_{2}}}=x$ for which solution freeze at

$-10^{o}C$

$\dfrac{V_{C_{2}H_{6}O_{2}}}{V_{Water}}=\dfrac{1}{x}\Rightarrow \dfrac{\dfrac{wt_{C_{2}H_{6}O_{2}}}{1.12}}{\dfrac{wt _{H_{2}O}}{1}}=\dfrac{1}{x}$

$\dfrac{wt_{C_{2}H_{6}O_{2}}}{1.12\times wt_{H_{2}O}}=\dfrac{1}{x}\Rightarrow \dfrac{wt_{C_{2}H_{6}O_{2}}}{wt_{H_{2}O}}=\dfrac{1.12}{x}$

Now,

$m=\dfrac{wt_{C_{2}H_{6}O_{2}}}{62\times wt_{H_{2}O}}\times 1000=\dfrac{1.12}{62x}\times 1000\quad \therefore m=\dfrac{1120}{62x}$

Now,

$\triangle T_{f}=K_{f}\times m\Rightarrow 1.86\times \dfrac{1120}{62x}=10\Rightarrow x=3.36$

A solution containing ethyl alcohol and propyl alcohol has a vapour pressure of $290\ mm$ at $30^{o}C$ . Find the vapour pressure of pure ethyl alcohol if its mole fraction in the solution is $0.65$ . The vapour pressure of propyl alcohol is $210\ mm$ at the same temperature.

$P_{solution}=290\ mm$ of

$Hg$

$P^{0}_{P.\ A}=210\ mm$ of

$Hg$

$X_{E.\ A}=0.65$

$\therefore X_{P.\ A}=1-X_{E.\ A}=1-0.65=0.35$

$\therefore P_{solution}=X_{E.\ A}.P^{0}_{E.\ A}+X_{P.\ A}.P^{0}_{P.\ A}$

$290=0.65\ P^{0}_{E.A}+0.35\times 210$

$P^{0}_{E.A}=\dfrac{290-73.5}{0.65}=333.12\ mm$ of

$Hg$

Benzene and toluene fomr nearly ideal solution. If at $27^{o}C$ the vapour pressures of pure toluene and pure benzene are $32.06\ mm$ and $103.01\ mm$ respectively.
Calculate the vapour pressure of a solution containing $0.60$ mole fraction of toluene.

According to the Raoult's Law:

$\mathrm{P_T = P^0_AX_A + P^0_BX_B}$

$\mathrm{P_T}$ for the given solution will be given by:

$\mathrm{P_T = 32.06 \times 0.60 + 103.01 \times (1-0.60)}$

$\mathrm{\implies P_T = 32.06 \times 0.60 + 103.01 \times0.40}$

$\mathrm{\implies P_T = 19.236 + 41.204}$ mm of Hg

$\mathrm{\implies P_T = 60.44}$ mm of Hg

So, thr vapor pressure of this solution will be 60.44 mm of Hg.

$10.03g$ of vinegar was diluted to $100mL$ and a $25mL$ sample was titrated with the $0.0176M$ $Ba{(OH)}_{2}$ solution $34.30mL$ was required for equivalence. What is the percentage of acetic acid in the vinegar?

$10.03g$ of vinegar (contain

${CH}_{3}COOH$ )

The solution is made upto

$100ml$ and out of which

$25ml$ is taken out for the experiment

$2{CH}_{3}COOH+Ba{(OH)}_{2}\rightarrow Ba{({CH}_{3}COO)}_{2}+2{H}_{2}O$

$V=25ml$

$0.0176 M$

$v.f=1$

$34.30ml$

$v.f=2$

$\cfrac { wt.\quad of\quad { CH }_{ 3 }COOH }{ eq.wt } \times 1000=34.30\times 0.0176\times 2=1.2074$

$\cfrac { wt.\quad of\quad { CH }_{ 3 }COOH }{ 60/1 } \times 1000=1.2074$

${CH}_{3}COOH$ in

$25ml=\cfrac{1.2074\times 60}{1000}=0.0724$

wt of

${CH}_{3}COOH$ in

$100ml=\cfrac{0.0724}{25}\times 100=0.2896$

% of

${CH}_{3}COOH$ in vinegar

$=\cfrac{0.289}{10.3}\times 100=2.80$ %

Igniting $MnO_{2}$ converts it quantitatively to $Mn_{3}O_{4}$ . A sample of pyrolusite is of the following composition: $MnO_{2}= 80\%; SiO_{2}$ and other inert constituents $15\%$ and rest being water.
The sample is ignited in air to constant weight. What is the percentage of $Mn$ in the ignited sample?

$[Mn=54.9, O=16]$

1714524_1730667_ans_2c5a6ec7db964fa598957f33b634a29c.jpg

An alloy of aluminium and copper was treated with aqueous $HCl$ The aluminium dissolved according to the reaction .
$\displaystyle Al + 3H^{+} \rightarrow Al^{+3} + \dfrac{3}{2}H_2 ,$
but the copper remained as pure metal . A $0.350-g$ sample of the alloy gave $415 cc$ of $H_2$ measured at $273 K$ and $1 \,atm$ pressure . What is the weight percentage of $Al$ in the alloy ?
Give your answer as nearest integer.

$Al + 3H^{+} \rightarrow Al^{+3} +\dfrac{3}{2}H_2$

Nobel metals

$( Ag , Au , Pt ,Hg , Cu )$ does not react with acid /

$H_2O$ to produce

$H_2$

Volume of

$H_2$ produced

$= 415 cm^{3}$

$\displaystyle \therefore \,$ moles of

$H_2$ produced

$= \dfrac{415}{22400}$

From balanced chemical reac

$^{-n}$

$\displaystyle \therefore \,\dfrac{3}{2}$ mole of

$H_2$ produced by

$1$ mole of

$Al$

$\displaystyle \therefore \, 1$ ---------------------------

$\dfrac{2}{3}$ ----------------

$\displaystyle \therefore \,\dfrac{415}{22400}$ -----------------------

$\dfrac{2}{3} \times \dfrac{415}{22400} = 0.01235$

$\displaystyle \therefore\,Wt.$ of

$Al = 0.01235 \times 27 \,g = 0.3335 \,g$

$\displaystyle \therefore\,$ % of

$Al = \dfrac{0.3335}{0.35} \times 100 = 95.3$ %
Ans-

$95$

Concentrated $HCl$ solution is $37.0$ % $HCl$ and has a density of $1.19g/mL$ . A dilute solution of $HCl$ is prepared by diluting $4.50mL$ of this concentrated $HCl$ solution to $100mL$ with water. Then $10mL$ of this dilute $HCl$ solution reacts with an $Ag{NO}_{3}$ solution. Calculate the volume of $0.108M$ $Ag{NO}_{3}$ solution required to precipitate all the chloride as $AgCl(s)$

Normality of concentrated

$HCl$ solution

$=\cfrac{37\times 1.19\times 10}{36.5}=12.06N$

By principle of dilution,

${V}_{1}{N}_{1}={V}_{f}{N}_{f}$

$4.5\times 12.06=100\times {N}_{f}$

${N}_{f}=0.543$

$HCl+Ag{NO}_{3}\rightarrow AgCl+H{NO}_{3}$

$10ml$

$0.108M$

$0.543N$

$V=$ ?

For complete precipitation,

meq of

$HCl=$ meq of

$Ag{NO}_{3}$

$10\times 0.543=0.108\times V_{Ag{NO}_{3}}$

$V_ {Ag{NO}_{3}}=50.27\ ml$

Twenty grams of HI is heated at $327^0C$ in a bulb of 1-litre capacity. Calculate the volume percentage of $H_2, \ I_2$ and HI at equilibrium. Given that the mass law constant for the equation $2HI \rightleftharpoons H_2 + I_2$ is 0.0559 at $327^0C$ when concentrations are expressed in moles / litre.

Initially,

$\underset{\dfrac{20}{128}M}{2HI (g)} \rightleftharpoons H_2(g) + I_2(g)$

$eq^m$

$\underset{\dfrac{20}{128} - 2\alpha M}{2HI (g)} \rightleftharpoons \underset{\alpha M}{H_2(g)} + \underset{\alpha M}{I_2(g)}$

Given, 20 gm of HI given at T = 600 K and 1 lit

At constant temperature & pressure,

Volume % = no. of moles %

$Keq = \dfrac{[H_2][I_2]}{[HI]^2} = \dfrac{\alpha^2}{\Big(\frac{20}{128} - 2\alpha \Big)^2} = 0.0559$

$\Rightarrow \dfrac{\alpha}{\frac{20}{128} - 2\alpha} = 0.236 \Rightarrow \alpha = \dfrac{0.037}{1.472} = 0.025$

Volume % of

$H_2= \dfrac{\alpha}{\frac{20}{128}}\times 100 = \dfrac{0.025}{20} \times 128 \times 100 = \dfrac{2.5 \times 128}{20} = 16 \%$

Volume % of

$I_2 = \dfrac{\alpha}{20/128} \times 100 = 16 \%$

Volume % of

$HI = 100 - (16 + 16)= 68 \%$

The vapour pressure of water is $3167.2$ Pa at $25^oC$ . What would be the vapour pressure of a solution of sucrose(with mole fraction of sucrose $=0.1$ ) and of a solution of levulose(with mole fraction of levulose $=0.1$ )?

We know that ,

$P_{soln}=P_{solvent}(1-x_{solute})$ .

Thus

$P_{sucrose} = P_{levulose} = 3167.2 \times 0.9 = 2850.5 \ Pa$

Liquid $A$ and $B$ form ideal solution over the entire range of composition. At temperature $T$ , equimolar binary solution of liquids $A$ and $B$ has vapour pressure $45\ torr$ . At the same temperature, a new solution of $A$ and $B$ having mole fraction $x_{A}$ and $x_{B}$ , respectively, has a vapour pressure of $22.5\ torr$ . The value of $x_{A}/x_{B}$ in the new solution is .....
$\mathrm{(P^0_A= 20 torr)}$

According to Raoult's Law:

$\mathrm{P_T =P^0_AX_A + P^0_BX_B}$

Case 1: Equimolar solution of A & B

$\mathrm{45 = P^0_A\times \cfrac{1}{2} + P^0_B\times \cfrac{1}{2}}$

$\mathrm{\implies P^0_A +P^0_B = 90 \ torr}$

$\mathrm{\implies P^0_B =90-P^0_A}$

$\mathrm{\implies P^0_B =70 \ torr}$

Case 2: Solution has a total vapor pressure of 22.5 torr

$\mathrm{22.5 = 20 \times X_A + 70 \times X_B}$

$\mathrm{X_A +X_B =1}$

$\mathrm{\implies X_B=1- X_B }$

$\mathrm{\implies 22.5 = 20 \times X_A + 70 \times (1-X_A)}$

$\mathrm{\implies X_A = 0.95}$

$\mathrm{\implies X_B = 0.05}$

So,

$\mathrm{\cfrac{X_A}{X_B} = \cfrac{0.95}{0.05}}$

$\mathrm{\implies \cfrac{X_A}{X_B} = 19}$

An aqueous solution freezes at $271.5$ K. Determine its boiling point and vapour pressure at $298$ K. The cryoscopic constant of water is $1.86^oC/m$ , its ebullioscopic constant is $0.516^oC/m$ and the water vapour pressure at $298$ K is $3168$ Pa.

1615387_1738261_ans_f4dd3c8d298e42a6a2dbe8f4b5a082aa.jpg

What weight of the nonvolatile solute, urea $(NH_2-CO-NH_2)$ needs to be dissolved in $100$ g of water in order to decrease the vapor pressure of water by $25\%$ ? What will be the molality of the solution?

2040945_1738673_ans_1b4a5e22dbd142deae1c7820f05feda9.jpg

What are the concentration and percentage of $Ag^+$ ion remaining after $Ag_2CrO_4$ precipitates when $25$ mL of $0.10$ M $AgNO_3$ is added to $25$ mL of $0.10$ M $K_2CrO_4$ ? $K_{sp}(Ag_2CrO_4)=1.1\times 10^{-12}$ .

1610647_1740108_ans_5d935075412d4a299d5a9a90d3779225.jpeg

The vapour pressure of pure water at $25^{o}C$ is $23.62\ mmHg$ . What will be the vapour pressure of a solution of $1.5\ g$ of urea in $50\ g$ is water?

$\dfrac{P^{0}_{Water}-P_{Solution}}{P^{0}_{Water}}=X_{Solute}$

$\dfrac{23.62-P_{solution}}{23.62}=\dfrac{\dfrac{1.5}{60}}{\dfrac{1.5}{60}+\dfrac{50}{18}}\quad (urea =CO(NH_{2})_{2}\ M.w=60)$

$=\dfrac{0.025}{2.80278}=8.92\times 10^{-3}$

$P_{solution}=23.62-(8.92\times 10^{-3})\times23.62=23.62-0.211=23.41\ mm$ of

$Hg$

The vapour pressure of a $0.01\ m$ solution of a weak base $BOH$ in water at $20^{o}C$ is $17.536\ mm$ . Calculate $K_{b}$ for the base. Aq. tension at $20^{o}C=17.54\ mm.$

Then apply $i=1+x$ and $K_{b}=\dfrac{0.01 x^{2}}{1-x}$

Relative lowering of vapour pressure

$=i X_{Solute}$

$\dfrac{17.54-17.536}{17.54}=iX_{Solute}$

$i=\dfrac{2.28\times 10^{-4}}{\dfrac{0.01}{0.01+1000/18}}=2.28\times 10^{-2}\times 55.56=1.267$

$BOH\rightleftharpoons B^{+}+ OH^{-}$

$1$

$0$

$1-x$

$x$

$\therefore i=\dfrac{1-x+x+x}{1}=1+x=1.267$

$x=1.267-1=0.267$

$\therefore K_{b}=\dfrac{0.01(x^{2})}{(1-x)}=\dfrac{(0.267)^{2}\times 10^{-2}}{(1-0.267)}=9.725\times 10^{-4}$

For dilute solution molality = molarity

The rate of diffusion of a sample of ozonised oxygen is $0.98$ times more than that of pure oxygen. Find the percentage (by volume) of ozone in the ozonised sample.

$3O_2 \rightleftarrows 2O_3$

Initially

$1$

$0$

At reacn

$1-3x$

$2x$

Total moles now

$=1-x$

$\therefore M.W.=\dfrac{32}{1-x}$

Now

$\dfrac{R_{O_2+O_3}}{R_{O_2}}=\sqrt{\dfrac{M_{O_2}}{M.W_{O_2+O_3}}}=\sqrt{\dfrac{32}{32/1-x}}$

$0.98=\sqrt{1-x}$

$1-x=0.96$

$\therefore x=0.04$

$\therefore$ Mole

$\%$ of ozone

$=\dfrac{2x}{1-x}\times 100=\dfrac{0.08}{0.96}\times 100=8.25\%$

$\therefore$ Volume

$\%$ = mole

$\%=8.33\%$

Assume liquefied petroleum gas (LPG) is a 50-50 (by mole) mixture of n -pentane and n-butane . calculate the calorific value (in kJ /mol) of gas available from a newly filled cylinder .give your answer divided by 100.

n-butane , $C_4 H_{10}$ n-pentane , $C_5 H_{12}$
Vapour pressure 1800 Torr 600 Torr
Calorific value 2800 kJ /mol 3600kJ /mol

First of all we will find mole fraction in liquid mixture , let mixture is 100 mole

So mole of pentane = 50

Moles of butane = 50

$X_{pentane } = 0.5 \quad \quad \quad X_{Butane} = 0.5$

$P_T = 600 \times 0.5 + 1800 \times 0.5 = 1200$

$Y_{pentane} = \dfrac {300}{1200} = \dfrac {1}{4} \quad \quad \quad \quad Y_{Butane} = \dfrac {900}{1200} = \dfrac { 3}{4}$

Calorific value =

$3600 \times \dfrac {1}{4} + 2800 \times \dfrac {3}{4} = 3000 kJ / mol$

During use

$Y_{pentane } \uparrow$ calorific value

$\uparrow$

Answer=

$\dfrac{3000}{100}=30$

A space capsule is filled with neon gas at $1$ atm and $290$ K. The gas effuses through a pinhole into outer space at such a rate that the pressure drops by $0.3$ mm/second. If the capsule were filled with $30\%$ He, $20\%$ $O_2$ and $50\%$ $N_2$ (mole $\%$ ) at a total pressure of $1$ atm and a temperature of $290$ K, calculate the rate of pressure drop.

From Graham's law of diffusion,

$-\dfrac{dP}{dt} = \dfrac{KP_o}{\sqrt{M}}$

$P_o = 1 \ atm = 760 \ mm , T = 290K, \ \dfrac{dP}{dt} = 0.3 \ mm/s, \ M_{Ne} = 20$

From the above equation,

$M_{NH_3}$ = 17

$-\dfrac{dP}{dt} = \dfrac{KP_o}{\sqrt{M}} =\dfrac{0.0017 \times 760}{\sqrt{17}} = 0.32 \ mm / sec$

Avg molecular mass =

$0.2 \times 4 + 0.2 \times 32 + 0.5 \times 28= 21.2$

$-\dfrac{dP}{dt} =\dfrac{KP_o}{\sqrt{M}} = \dfrac{0.0017 \times 760}{\sqrt{21.2}} = 0.29 \ mm / sec$

A solution A (l) and B (l) with 30 mole percent of A is in equilibrium with its vapour which contains 60 mole percent of A . Assuming ideality of the solution and its vapour calculate the ratio of vapour pressure of pure A to that of pure B. (report your answer as ratio $\times$ 2 )

In Solution ,

$x_A = 0.3 ; x_B = 0.7$

In vapour phase ,

$x_A = 0.6 ; x_B = 0.4$

$x'_A = \dfrac {P_A}{P} = 0.6$

$\dfrac {P_A}{P_A + P_B} = \dfrac {p^0_Ax_A }{p^0_Ax_A + p^0_Bx_B }$

$0.6 = \dfrac {0.3 p^0_A}{0.3 p^0_A + 0.7 p^0_B } ..........(1)$

$x'_B = \dfrac {P_B}{P} = \dfrac {P_B}{P_A + P_B} = 0.4$

$\Rightarrow \dfrac { 0.7 p^0_B}{ 0.3 p^0_A + 0.7 p^0_B} = 0.4$ .............(2)

Dividing (1) by (2)

$\dfrac {0.3 P^0_A}{0.7p^0_B} = \dfrac {0.6}{0.4 }$

$\Rightarrow \dfrac {p^0_A}{p^0_B} = \dfrac { 0.6 \times 0.7}{0.4 \times 0.3} = \dfrac {7}{2} = 3.5$

Calculate the freezing point depression expected for $0.0711 \, m$ aqueous solution of $Na_2SO_4$ . If this solution actually freezes at $-0.320^{\circ}\,C$ , what would be the value of van't Hoff factor?
( $K_f$ for water is $1.86 \,K \, kg \, mol^{-1}$ )

$\Delta T_f = [273.15 - ( -0.320 + 273 .15)]K = 0.320 \,K$

$\Delta T_f = K_f . m = 1.86 \,K kg \, mol^{-1} \times 0.0711 \, mol \,kg^{-1} = 0.132\,K$

$i = \dfrac{{\text{Observed value of }}\Delta T_f}{{\text{Calculated value of }}\Delta T_f} = \dfrac{0.320\,K}{0.132 \,K} = 2.42$

What do you mean by $10$ % aqueous solution of sodium carbonate?

It means that

$10 \,g$ of

$Na_2CO_3$ is present in

$100 \,g$ of the solution

Define van't Hoff factor.

Van't Hoff factor may be defined as the ratio of normal molecular mass to observed molecular mass or the ratio of observed colligative property to normal colligative property.

Define the following terms :
Van't Hoff factor (i)

Van't Hoff factor (i) gives the extent of association or dissociation of the solute particles in the solution. It may be defined as the ratio of observed colligative property to calculated colligative property.

$i = \dfrac{{\text{Observed colligative property}}}{{\text{Calculated colligative property}}}$

What do you understand by 'colligative properties'?

Colligative properties are those properties which depend upon the number of moles of the solute particle but not on the nature of solute particles.

What is the van't Hoff factor for a compound which undergo dimerisation in an organic solvent?

$2X \rightarrow X_2$

$i = \dfrac{{\text{Number of moles of particles after association}}}{{\text{Number of moles of particles before association}}} = \dfrac{1}{2}$

What is collodion?

Collodion is a

$4\%$ solution of nitrocellulose in a mixture of alcohol and ether.

The dissolution of ammonia chloride in water is an endothermic process but still dissolves in water readily. Why?

This is because of the entropy change. In case case,

$\Delta S$ is +ve.

$NH_4Cl(aq) \rightarrow NH^{+}_{4} (aq) + Cl^-(aq)$

The ions that were held together in crystalline solid are free and moving in all possible directions. Its entropy has increased and this makes

$T \Delta S > \Delta H$ , i.e.,

$\Delta G = -ve$ .

Give Reasons for the Following:
Why is it said that solubility of any solute changes with a change in temperature?

Solvent dissolves solute particles by overcoming the intermolecular forces between the solute (re one can call it the solute-solute interaction). So, to dissolve a solute, the solute-solute interaction must be broken by the solvent particles.
When the temperature increases, the kinetic energy of the solvent particles increases, causing to break the solute-solute interactions more effectively. This allows the addition of more solute particles, as there Is space to accommodate more solute particles. So the solubility Increases with Increase In temperature. Lowering the temperature does the opposite. It decreases the solubility of the solute.

Which of the liquids in each of the following pair has higher vapour pressure: Mercury, water

$Water$

Water has a weaker inter-particle force. Hence, Water has a higher vapour pressure than mercury.

How many types of solutions are formed? Write briefly about each type with an example.

In a solution, the constituent particle which is in a larger amount is called a solvent and that in smaller quantities is called a solute.

Since the solvent and solute may be either gaseous, liquid and solid, the number of possible types of binary solutions than can be prepared are given below.

Solute	Solvent	Example
Gas	Gas	Air
Liquid	Gas	Humidity in air
Solid	Gas	Vapours of Iodine in air
Gas	Liquid	Aerated water ( $CO_2$ in $H_2O$ )
Liquid	Liquid	Alcohol in Water
Solid	Liquid	Sugar in water
Gas	Solid	$H_2$ in Palladium
Liquid	Solid	Amalgam of Mercury with $Na$
Solid	Solid	Copper in Gold (Alloys)

Which of the liquids in each of the following pair has higher vapour pressure : Petrol, kerosene

Petrol will have high vapour pressure than kerosene as petrol will produce vapors more easily than kerosene at the same temperature.

Give reason for the following :
Blue solution of copper sulphate changes to green when a piece of iron is added to this solution.

$\text{This is because the blue colour of the copper sulphate solution fades}$ and

$\text{eventually turns into light green due to the formation of ferrous sulphate}$ .

A solution of A and B with 30 mole percent of A is in equilibrium with its vapour which contain 60 mole percent of A. Assuming that the solution and the vapour behave ideally, calculated the ratio of the vapour pressures of pure A and pure B.

In solution

$x_A=0.30$

$\therefore x_B=0.70$
in the vapour phase,

$y_A=0.60$

$\therefore y_B=0.40$

But

$y_A=\dfrac{p_A}{Total\,V.P}=\dfrac{p_A}{p_A+p_B}=\dfrac{x_Ap^0_A}{x_Ap^0_A+x_Bp^0_B}=\dfrac{0.30p^0_A}{0.30p^0_A+0.70p^0_B}=0.60$ .........(i)

$y_B=\dfrac{p_B}{p_A+p_B}=\dfrac{x_Bp^0_B}{x_Ap^0_A+x_Bp^0_B}=\dfrac{0.70p^0_B}{0.30p^0_A+0.70p^0_B}=0.40$ ...........(ii)

Dividing eqn (1) by eqn (ii)

$\dfrac{0.30p^0_A}{0.70p^0_B}=\dfrac{0.60}{0.40}$ or

$\dfrac{p^0_A}{p^0_B}=\dfrac{0.60}{0.40}\times \dfrac{0.70}{0.30}=3.5$

(a) Benzoic acid completely dimerises in benzene. What will be the vapour pressure of a solution containing 61g of benzoic acid per 500g benzene when the vapour pressure of pure benzene at the temperature of the experiment is 66.6 torr?
(b) What would have been the vapour pressure is the absence of dimerisation?

(i)

$2C_6H_5COOH\rightarrow (C_6H_5COOH)_2$
Initial

$1$ mole

$0$
After assoc.

$0$

$\dfrac{1}{2}$ mole

$\therefore i=\dfrac{1}{2}$

$\dfrac{p^0-p_s}{p^0}=i\dfrac{n_2}{n_1+n_2}=\dfrac{1}{2}\times \dfrac{61/122}{61/122+500/78}=\dfrac{1}{2}\times \dfrac{0.5}{0.5+6.4}=0.036$

or

$1-\dfrac{p_s}{p^0}=0.036$ or

$\dfrac{p_s}{p^0}=0.964$ or

$p_s=0.964\times 66.6torr=64.2\,torr$

(ii) In the absence of dimerisation ,

$\dfrac{p^0-p_s}{p^0}=\dfrac{n_2}{n_1+n_2}=0.072$ or

$1-\dfrac{p_s}{p_0}=0.072$

or

$\dfrac{p_s}{p_0}=0.928$ or

$p_s=0.928\times 66.6\,torr=61.8\,torr$

Define the terms 'Vapour pressure and 'Solubility'.

$\text { Vapour Pressure of any liquid at any temperature } \\$

$\text { is the pressure exerted by vapours present } \\$

$\text { above liquid in equilibrium with liquid at that } \\$

$\text { temperature. } \\$

$\text { The maximum amount of solute that can be } \\$

$\text { dissolved in given quantity of solvent at a } \\$

$\text { given temp is called solubility. }$

Column-I	Column-II
(a) Kinetic energy	(p) Mole fraction
(b) Partial pressure of a gas	(q) Density
(c) Rate of diffusion	(r) Molar mass
(d) Vapour pressure of a liquid	(s) Absolute temperature

	n-butane , $C_4 H_{10}$	n-pentane , $C_5 H_{12}$
Vapour pressure	1800 Torr	600 Torr
Calorific value	2800 kJ /mol	3600kJ /mol

Solutions - Class 12 Engineering Chemistry - Extra Questions

The mass percentage of nitrogen in histamine is _______.

An ideal aqueous solution containing liquid A(M.Wt=128)A(M.Wt=128) 6464% by weight has a vapour pressure of 145mm145mm HgHg. If the vapour pressure of AA is xx mmmm of HgHg and that of water is 155mm155mm HgHg at the same temperature. Then find x5\dfrac{x}{5}. The solutions is ideal.

Does a new substance form when a solute dissolves in a solvent?

Calculate the percentage of magnesium in magnesium carbonate.

Mention names of four methods of expressing concentration of a solution.

Calculate percentage of oxygen present in CO2CO_{2}. [Atomic mass of O=16u. C=12u]

At constant temperature and pressure, one mole of a pure substance free energy is identical with:

Van't Hoff factor for an electrolyte is greater than _____.

For strong electrolytes, van't Hoff factor 'i' equals to total number of _____ (ions/atoms) in the formula unit.

Van't Hoff factor for TH(NO3)4TH(NO_3)_4 in an aqueous solution is :

Match column I with column II

Match column I with column II

For weak electrolytes, van't Hoff factor 'i' is a measure of the degree of dissociation of weak electrolyte.If true enter 1, if false enter 0.

Van't Hoff factor of a mixture of two moles of KI with 1 mole HgI2HgI_2 in a solution of water is :

What is the difference between lyophobic sol and lyophilic sol ?

How is the vapour pressure of a solvent affected when a non-volatile solute is dissolved in it ?

For an ideal solution, ΔmixH{\Delta}_{mix}H is _______ .

For an ideal solution, ΔmixV{\Delta }_{mix}V is

The vapour pressure of water is 12.3 kPa12.3\ kPa at 300 K300\ K. Calculate the vapour pressure of 11 molal solution of a non-volatile solute in it.

Vapour pressure of water at 293 K293\ K is 17.535 mm Hg17.535\ mm\ Hg. Calculate the vapour pressure of water at 293 K293\ K when 25 g25\ g of glucose is dissolved in 450 g450\ g of water.

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CNKCl,\ CH_{3}OH,\ CH_{3}CN

Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Define an ideal solution and write one of its characteristics.

Arrange the following compounds in increasing order of solubility in water: C6H5NH2,(C2H5)2NH,C2H5NH2{ C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2 },{ \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }NH,{ C }_{ 2 }{ H }_{ 5 }N{ H }_{ 2 }

Vapour pressure of three liquids A, B, C are in their ascending order. Identify the liquid that has least critical temperature in its gaseous state.

Why is be solubility of NaClNaCl in water 311gl−1311\,g\, l^{-1} NN but is zero in benzene?

What is the value of ΔHmix\Delta H_{mix} for an ideal solution ?

a. Which gas is dissolved in soft drinks?b. What will you do to increase the solubility of this gas ?

What happens to vapour pressure of water if a tablespoon of sugar is added to it?

What will be the value of van't Hoff factor (i)\left(i\right) of benzoic acid if it demerises in aqueous solution? How will the experimental molecular weight vary as compared to the normal molecular weight?

What will be the value of Van't Hoff factor for ethanoic acid in benzene?

Derive van't Hoff general solution equation.

What is relative lowering of vapour pressure? How is it useful to determine the molar-mass of a solute?

The vapour pressure of pure benzene at a certain temperature is 0.8500.850 bar. A non-volatile, non-electrolyte solid weighing 0.50.5 gg when added to 3939 gg of benzene (molar mass 7878 gg mol−1{mol}^{-1}), vapour pressure becomes 0.8450.845 bar. What is the molar mass of the solid substance?

'The extent to which a solute is dissociated or associated can be expressed by Van 't Hoff factor'. Substantiate the statement.

Derive the relationship between relative lowering of vapour pressure and molar mass of solute.

Alkaline solution of HgCl2Hg{Cl}_{2} and KIKI is called _________.

Mention the enthalpy of mixing (△mixH)({\triangle}_{mix}H) value to form an ideal solution.

Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquids are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure, in bar, of a mixture of 25 g heptane and 35 g of octane?

An aqueous solution containing 28%28\% by mass of a liquid A (mol. mass = 140140) has a vapour pressure of 160160 mm of 37oC37^o C. Find the vapour pressure of the pure liquid A (The vapour pressure of water at 37oC37^oCis 150150 mm).

Potassium ferrocyanide is 50% ionised in aqueous solution. Its van't Hoff factor will be:

A solution containing 8.44g8.44 g of sucrose in 100g100 g of water has a vapour pressure 4.56mm4.56 mm of HgHg at 273K273 K. If the vapour pressure of pure water is 4.58mm4.58 mm of HgHg at the same temperature, calculate the molecular weight of sucrose.

Calculate the vapour pressure of an aqueous solution of 1.0 molal glucose solution at 100oC.100^oC.

The freezing point of a solution containing 5.85 g of NaClNaCl in 100g of water is −3.348o-3.348^oC. Calculate van't Hoff factor 'i' for this solution. What will be the experimental molecular weight of NaCI?(KfK_f for water = 1.86Kkgmol−11.86 K kg mol^{-1}, at. wt. Na = 23, Cl = 35.5)

Two moles of O2O_2 gas is collected over water at 400 K temperature in 2 litre vessel. If the pressure of dry O2O_2 gas is 32.20 bar then find the vapour pressure of water under the same conditions.

The vapour pressure of a pure liquid at 2525 is 100 mm HgHg. Calculate the relative lowering of vapour pressure if the mole fraction of solvent in solution is 0.8.

Identify the solute & the solvent in the following solution:a) sugar solutionb) Airc) Rain waterd) Aerated drinks

Define an ideal solution.

At 90o90^oC, the vapour pressure of toluene is 400400 Torr and that of o-xylene is 150150 Torr. What is the composition of the liquid mixture that boils at 90o90^oC when the pressure is 0.500.50 atm? What is the composition of the vapour produced?

Calculate the following.The vapour pressure of the solution at 298298 K.

At 25o25^oC the saturated vapour pressure of water is 3.1653.165 kPa (23.7523.75 mm Hg). Find the saturated vapour pressure of a 5%5\% aqueous solution of urea (carbamide) at the same temperature. (Molar mass of urea =60.05=60.05 g mol−1mol^{-1})

The vapour pressure of an aqueous solution of glucose is 750mm at 373K. find molality and mole fraction

Identify the solute & the solvent in the following solutions:(a) Sugar solution (b) Air (c) Grains (d) Aerated drinks.

Derive the relationship between relative lowering of vapour pressure and mole fraction of the volatile liquid.

Out of two 0.10.1 molal solutions of glucose and of potassium chloride, which one will have a higher boiling point and why?

400 g400\ g of 20% (w/w)20\%\ (w/w) solution is cooled. 50 g50\ g of solute is precipitated. What is the %(w/w) of solute in the remaining solution?

Vapour pressure of liquids AA and BB at 298 K298\ K is 300 mm Hg300\ mm\ Hg and 450 mm Hg450\ mm\ Hg. Calculate the mole fraction of AA in the mixture. [Given total vapour pressure of solution=405 mm Hg=405\ mm\ Hg.]

The vapor pressure of water is 12.3 KPa12.3 \text { }KPa at 300K300K. Calculate the vapor pressure of 11 molal solution of a non-volatile solute in it.

What is vapour pressure?

The vapour pressure of water is 9292mm at 323323K. 18.118.1g of urea are dissolved in 100100g of water. The vapour pressure is reduced by 55mm. Calculate the molar mass of urea.

If equal volume of BaCl2{ BaCl }_{ 2 } and NaFNaF solutions are mixed , which of these combination will give a precipitate? Ksp{ K }_{ sp } of BaF2=1.7×10−7. { BaF}_{ 2} ={ 1.7\times 10 }^{ -7 }.

The vapour pressure of a pure liquid A is 8080 mmHg at 300K300K. It forms an ideal solution with liquid B. When the mole fraction of B is 0.40.4, the total pressure was to be 8888mmHg. The vapour pressure of liquid B would be?

The vapour pressure of solution of 5g of non electrolyte in 100 g of a water at a particular temperature is 2985Nm−22985 N{ m }^{ -2 }. The vapour pressure of pure water at that temperature is 3000Nm−23000 N{ m }^{ -2 }

What is the mass of a non volatile solute (molecular mass 60) that needs tobe dissolved in 100g 100 g of water In order to decrease the vapour pressure of water by 25 25% .What will be the velocity of solution?

If p° and p are the vapour pressure of a solvent and it's solution repetitively and N1 and N2 are the mole fractions of the solvent and solute respectively, then the relation between them is?

The vapour pressure of pure water is 23.8 mm Hg at 250C25^0C What is the vapour pressure of 2.50 molal C6H12O6C_6H_{12}O_6.

2 g benzoic acid (C6H5COOH)(C_{6}H_{5}COOH) dissolved in 25 g of benzene shoes a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol−1mol^{-1}. What is the percentage association of acid if it forms dimer in solution?

Explain homogenous and heterogenous conditions with an example.

Derive van"t Hoff equation for osmotic pressure of a solution.

A sample of O2O_{2} gas is collected over at 23o C23^{o}\ C at a barometric pressure of 751 mm Hg751\ mm\ Hg (vapour pressure of water at 23o C23^{o}\ C is 21 mmHg)21\ mmHg). The partial pressure of O2O_{2} gas in the sample collected is

A 0.2%0.2\% aq. sol. of a non-volatile solution exerted a VP of 1.0041.004 bar at 100∘C100^\circ C. What is the molar mass of the solute ? (Given V.P of mass water at 100∘C100^\circ C is 1.0131.013 bar and molar mass of water is 18g/mol18\,g/mol).

A 821 mL N2(g) N_2(g) was collected over liquid water at 300 K and 1 atm. If vapour pressure of H2O H_2O is 30 torr then moles of N2(g) N_2(g) in moist gas mixture is :

Define Van't Hoff's factor.

The vapour of a sodium containing 13×10−3 kg13\times 10^{-3}\ kg of solute in 0.1 kg0.1\ kg of water at 298 K298\ K is 27.371 mm Hg27.371\ mm\ Hg. Calculate the molar mass of the solute. Given that the vapour pressure of water at 298 K298\ K is 28.065 mm Hg28.065\ mm\ Hg .

Find its mass of urea dissolved in 100gH2O100g H_2O which decreese the vapour presure of solvent by 20%20\%

110g of salt is present in 550g of solution. Calculate the mass percentage of the solution.

The vapour pressure of pure water at is 25.21 torr. What is the vapour pressure of a solution which contain 20.0 glucose, C6H12O6{C_6}{H_{{\text{12}}}}{{\text{O}}_6} in 70 g water ?

Two A and B in a mixture produced 550 mm of Hg vapour pressure. If their vapour pressure in the pure states are 400 mm of Hg and 600 mm of Hg respectively. What is the vapour pressure of 'A' in the mixture?

Calculate the percentage by mass to volume for 0.2 M NaOH solution.

An ideal aqueous solution containing liquid $A(M.Wt=128)$ $64$ % by weight has a vapour pressure of $145mm$ $Hg$ . If the vapour pressure of $A$ is $x$ $mm$ of $Hg$ and that of water is $155mm$ $Hg$ at the same temperature. Then find $\dfrac{x}{5}$ . The solutions is ideal.

Calculate percentage of oxygen present in $CO_{2}$ . [Atomic mass of O=16u. C=12u]

Van't Hoff factor for $TH(NO_3)_4$ in an aqueous solution is :

For weak electrolytes, van't Hoff factor 'i' is a measure of the degree of dissociation of weak electrolyte.
If true enter 1, if false enter 0.

Van't Hoff factor of a mixture of two moles of KI with 1 mole $HgI_2$ in a solution of water is :

For an ideal solution, ${\Delta}_{mix}H$ is _______ .

For an ideal solution, ${\Delta }_{mix}V$ is

The vapour pressure of water is $12.3\ kPa$ at $300\ K$ . Calculate the vapour pressure of $1$ molal solution of a non-volatile solute in it.

Vapour pressure of water at $293\ K$ is $17.535\ mm\ Hg$ . Calculate the vapour pressure of water at $293\ K$ when $25\ g$ of glucose is dissolved in $450\ g$ of water.

Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain.
Cyclohexane, $KCl,\ CH_{3}OH,\ CH_{3}CN$

Arrange the following compounds in increasing order of solubility in water:
${ C }_{ 6 }{ H }_{ 5 }N{ H }_{ 2 },{ \left( { C }_{ 2 }{ H }_{ 5 } \right) }_{ 2 }NH,{ C }_{ 2 }{ H }_{ 5 }N{ H }_{ 2 }$

Why is be solubility of $NaCl$ in water $311\,g\, l^{-1}$ $N$ but is zero in benzene?

What is the value of $\Delta H_{mix}$ for an ideal solution ?

a. Which gas is dissolved in soft drinks?
b. What will you do to increase the solubility of this gas ?

What will be the value of van't Hoff factor $\left(i\right)$ of benzoic acid if it demerises in aqueous solution? How will the experimental molecular weight vary as compared to the normal molecular weight?

The vapour pressure of pure benzene at a certain temperature is $0.850$ bar. A non-volatile, non-electrolyte solid weighing $0.5$ $g$ when added to $39$ $g$ of benzene (molar mass $78$ $g$ ${mol}^{-1}$ ), vapour pressure becomes $0.845$ bar. What is the molar mass of the solid substance?

Alkaline solution of $Hg{Cl}_{2}$ and $KI$ is called _________.

Mention the enthalpy of mixing $({\triangle}_{mix}H)$ value to form an ideal solution.

An aqueous solution containing $28\%$ by mass of a liquid A (mol. mass = $140$ ) has a vapour pressure of $160$ mm of $37^o C$ . Find the vapour pressure of the pure liquid A (The vapour pressure of water at $37^oC$ is $150$ mm).

A solution containing $8.44 g$ of sucrose in $100 g$ of water has a vapour pressure $4.56 mm$ of $Hg$ at $273 K$ . If the vapour pressure of pure water is $4.58 mm$ of $Hg$ at the same temperature, calculate the molecular weight of sucrose.

Calculate the vapour pressure of an aqueous solution of 1.0 molal glucose solution at $100^oC.$

The freezing point of a solution containing 5.85 g of $NaCl$ in 100g of water is $-3.348^o$ C. Calculate van't Hoff factor 'i' for this solution. What will be the experimental molecular weight of NaCI?
( $K_f$ for water = $1.86 K kg mol^{-1}$ , at. wt. Na = 23, Cl = 35.5)

Two moles of $O_2$ gas is collected over water at 400 K temperature in 2 litre vessel. If the pressure of dry $O_2$ gas is 32.20 bar then find the vapour pressure of water under the same conditions.

The vapour pressure of a pure liquid at $25$ is 100 mm $Hg$ . Calculate the relative lowering of vapour pressure if the mole fraction of solvent in solution is 0.8.

Identify the solute & the solvent in the following solution:
a) sugar solution
b) Air
c) Rain water
d) Aerated drinks

At $90^o$ C, the vapour pressure of toluene is $400$ Torr and that of o-xylene is $150$ Torr. What is the composition of the liquid mixture that boils at $90^o$ C when the pressure is $0.50$ atm? What is the composition of the vapour produced?

Calculate the following.
The vapour pressure of the solution at $298$ K.

At $25^o$ C the saturated vapour pressure of water is $3.165$ kPa ( $23.75$ mm Hg). Find the saturated vapour pressure of a $5\%$ aqueous solution of urea (carbamide) at the same temperature. (Molar mass of urea $=60.05$ g $mol^{-1}$ )

Identify the solute & the solvent in the following solutions:
(a) Sugar solution (b) Air (c) Grains (d) Aerated drinks.

Out of two $0.1$ molal solutions of glucose and of potassium chloride, which one will have a higher boiling point and why?

$400\ g$ of $20\%\ (w/w)$ solution is cooled. $50\ g$ of solute is precipitated. What is the %(w/w) of solute in the remaining solution?

Vapour pressure of liquids $A$ and $B$ at $298\ K$ is $300\ mm\ Hg$ and $450\ mm\ Hg$ . Calculate the mole fraction of $A$ in the mixture.

[Given total vapour pressure of solution $=405\ mm\ Hg$ .]

The vapor pressure of water is $12.3 \text { }KPa$ at $300K$ . Calculate the vapor pressure of $1$ molal solution of a non-volatile solute in it.

The vapour pressure of water is $92$ mm at $323$ K. $18.1$ g of urea are dissolved in $100$ g of water. The vapour pressure is reduced by $5$ mm. Calculate the molar mass of urea.

If equal volume of ${ BaCl }_{ 2 }$ and $NaF$ solutions are mixed , which of these combination will give a precipitate?
${ K }_{ sp }$ of ${ BaF}_{ 2} ={ 1.7\times 10 }^{ -7 }.$

The vapour pressure of a pure liquid A is $80$ mmHg at $300K$ . It forms an ideal solution with liquid B. When the mole fraction of B is $0.4$ , the total pressure was to be $88$ mmHg. The vapour pressure of liquid B would be?

The vapour pressure of solution of 5g of non electrolyte in 100 g of a water at a particular temperature is $2985 N{ m }^{ -2 }$ . The vapour pressure of pure water at that temperature is $3000 N{ m }^{ -2 }$

What is the mass of a non volatile solute (molecular mass 60) that needs tobe dissolved in $100 g$ of water In order to decrease the vapour pressure of water by $25%$ .What will be the velocity of solution?

The vapour pressure of pure water is 23.8 mm Hg at $25^0C$ What is the vapour pressure of 2.50 molal $C_6H_{12}O_6$ .

2 g benzoic acid $(C_{6}H_{5}COOH)$ dissolved in 25 g of benzene shoes a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg $mol^{-1}$ . What is the percentage association of acid if it forms dimer in solution?

A sample of $O_{2}$ gas is collected over at $23^{o}\ C$ at a barometric pressure of $751\ mm\ Hg$ (vapour pressure of water at $23^{o}\ C$ is $21\ mmHg)$ . The partial pressure of $O_{2}$ gas in the sample collected is

A $0.2\%$ aq. sol. of a non-volatile solution exerted a VP of $1.004$ bar at $100^\circ C$ . What is the molar mass of the solute ?
(Given V.P of mass water at $100^\circ C$ is $1.013$ bar and molar mass of water is $18\,g/mol$ ).

A 821 mL $N_2(g)$ was collected over liquid water at 300 K and 1 atm. If vapour pressure of $H_2O$ is 30 torr then moles of $N_2(g)$ in moist gas mixture is :

The vapour of a sodium containing $13\times 10^{-3}\ kg$ of solute in $0.1\ kg$ of water at $298\ K$ is $27.371\ mm\ Hg$ . Calculate the molar mass of the solute. Given that the vapour pressure of water at $298\ K$ is $28.065\ mm\ Hg$ .

Find its mass of urea dissolved in $100g H_2O$ which decreese the vapour presure of solvent by $20\%$

The vapour pressure of pure water at is 25.21 torr. What is the vapour pressure of a solution which contain 20.0 glucose, ${C_6}{H_{{\text{12}}}}{{\text{O}}_6}$ in 70 g water ?

When $HgI_2$ is added to aqueous solution of KI, why there is an increase in vapour pressure of solution ?

Predict whether van't HOff factor $(i)$ is less than one or greater than one in the following:
(i) $CH_3COOH$ dissolved in water
(ii) $CH_3COOH$ dissolved in benzene

The vapour pressure of water at $20^oC$ is $17$ mm Hg.Calculate the vapour pressure of a solution containing $2.8$ g of urea $(NH_2CONH_2)$ in $50$ g of water ( $16.71$ mm Hg).

A blackened silver coin weighing $15\ g$ on treatment with $HCl$ yielded $25\ mL$ of $H_{2}S$ at $12^{o}C$ and $775\ mm$ pressure. What percentage of original silver tarnished?

$3\ g$ sample of blue vitrol $(CuSO_{4}.5H_{2}O)$ were dissolved in water. $BaCl_{2}$ solution was mixed in excess to this solution. The precipitate obtained was washed and dried. It weighed $2.82\ g$ . Determine $\%$ of $SO_{4}^{2-}$ by weight in sample.

An aqueous solution containing $20\%$ by weight of liquid $X(mol. wt. =140)$ has a vapour pressure $160\ mm$ at $60^{o}C$ . Calculate the vapour pressure of pure liquid $X$ if the vapour pressure of water is $150\ mm$ at $60^{o}C$

Which of the following aqueous solutions has a higher vapour pressure if the density of water is $1\ g/cc$ ?
(1) Solution having mole fraction of cane sugar $=0.1$
(2) Solution having molal concentration = 1 m
[Give your answer in terms of 1 or 2, e.g. enter 1 if (1) is true.]

Which of the following aqueous solutions has a higher vapour pressure if the density of water is $1\ g/cc$ ?
Solution having molal concentration $=1\ m$

What is the vapour pressure at $100^{o}C$ of a solution containing $15.6\ g$ of water and $1.68\ g$ of sucrose $(C_{12}H_{22}O_{11})$ ?

One litre of water was added to $500\ mL$ of $32\%\ HNO_{3}$ of density $1.20\ g/mL$ . What is the per cent concentration of $HNO_{3}$ in the solution obtained?

$0.5\ g$ of a nonvolatile organic compound $(mol.wt. 65)$ is dissolved in $100\ mL$ of $CCl_{4}$ . If the vapour pressure pure $CCl_{4}$ is $143\ mm$ , what would be the vapour pressure of the solution? (Density of $CCl_{4}$ solution is $1.58\ g/cc$ )

Determine the per cent concentration of a solution obtained by mixing $300\ g$ of a $25\%$ and $400\ g$ of a $40\%$ solution.

When $10$ mL of ethanol of density $0.7893$ g/L is mixed with $20$ mL of water of density $0.9971$ g/L at $25^o$ C, the final solution has a density of $0.9571$ g/L. Calculate the percentage change in total volume on mixing.

Calculate the concentration of a solution obtained by mixing $300\,g \,25$ % by weight solution of $NH_4Cl$ and $150\,g$ of $40$ % by weight solution of $NH_4Cl.$

If the vapour pressure of pure liquids A and B are $300$ mm and $800$ mm Hg at $75^o$ C, calculate the composition of the mixture such that it boils at $75^o$ C. Find the composition of the vapour phase.

Calculate the % of free $SO_3$ in oleum (a solution of $SO_3$ in $H_2SO_4$ ) that is labelled $109$ % $H_2SO_4$ by weight.

A $0.5$ g sample containing $MnO_2$ is treated with HCl, liberating $Cl_2$ . The $Cl_2$ is passed into a solution of KI and $30.0 cm^3$ of $0.1$ M $Na_2S_2O_3$ are required to titrate the liberated iodine. Calculate the percentage of $MnO_2$ in the sample.

Both Sarika and Mohan were asked to make a salt solution. Sarika was given a teaspoonful of salt and half a glass of water, whereas Mohan was given twenty teaspoons full of salt and half a glass of water.
(a) How would they make salt solutions?
(b) Who would be able to prepare a saturated solution?

Differentiate between the following.
Solution and Suspension

Calculate the mass of $95\%$ pure $MnO_2$ to produce $35.5 \,g$ of $Cl_2$ as per the following reaction:
$MnO_2 + 4HCl \rightarrow {}MnCl_2 + Cl_2 + H_2O.$ (At. mass of $Mn = 55$ )

A solution of sodium hydroxide contains $4.8 \,g$ of the substance per $dm^3,$
What volume of water has to be added to one $dm^3$ of the solution in order to make it exactly decinormal ?

Calculate the volume of concentrated nitric acid of normality $14$ required to prepared $1 \,dm^3$ of $\dfrac{N}{10}$ nitric acid.

Calculate the percentage of nitrogen in $NH_4.$ (Atomic mass of $N = 14, H = 1$ amu)

Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid (iv) ethylene glycol (v) chloroform (vi) pentanol