Class 11 Medical Chemistry - Some Basic Concepts Of Chemistry

Calculate the concentration of NaOH solution in g/mL, which has the same normality as that of a solution of HCl of concentration 0.04 g/mL.

$N_{HCl}=\dfrac{W_B\times 1000}{E_B\times V}=\dfrac{0.04\times 1000}{36.5 \times 1}=1.095$

$N_{NaOH}=N_{HCl}$

$\therefore 1.095=\dfrac{W_B\times 1000}{40\times 1}$

$W_B = 0.0438 g/mL$

Silver crystallises in fcc lattice. If edge length of the cell is $4.07\times {10}^{-8}cm$ and density is $10.5g$ ${cm}^{3}$ , calculate the atomic mass of silver.

Given, edge length (l)=

$4.07 \times 10^{-8}$ cm

density (d)=10.5 g/cc

Number of atoms in fcc lattice=4

Avagadro's number=

$6.023 \times 10^{23}$

Density

$=d=\cfrac { ZM }{ { l }^{ 3 }{ N }_{ A } }$

$\Longrightarrow M=\cfrac { d{ l }^{ 3 }{ N }_{ A } }{ Z } =\cfrac { (10.5){ (4.07) }^{ 3 }(6.023\times 10^{ 23 }) }{ 4 } =107.09g{ mol }^{ -1 }$

$H_2SO_4$ is 98% by weight of solution, $M x 10^2$ . What is the value of M?

$98$ % by weight means

$98gm$

$H_2SO_4$ in

$2g$ water.

Mass of solution=

$100= 1\times 10^2$

$M=1$ .

Insulin contains $3.4$ % sulpur calculate minimum molecular weight of insulin.

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1 atmosphere means how many Pascal?

$1atm=1.01325bar\\ 1bar={ 10 }^{ 5 }pascal\\ \Longrightarrow 1atm=1.01325\times { 10 }^{ 5 }pascal\\ i.e.,\quad 1atm=101325pascal\\ Pascal\quad is\quad the\quad S.I.\quad unit\quad of\quad pressure.$

Which has the highest mass ?
(a) 50 g iron (b) 5 moles of ${ N }_{ 2 }$
(c) 0.1 mol atom of Ag (d) ${ 10 }^{ 23 }$ atoms of carbon

the correct option is b

We shall convert all values into grams

a.) 50 g fe weigh=50g

b.)5 moles of

$N_2$ =

we know no. of moles=

$\frac{Weights of substance}{molecular mass}$

$5=\frac{Weight of N_2}{28}$

weight of

$N_2$ = 140 g

c.)

$0.1 g atom of Ag=0.1g$

d.)

$10^{23} atom of carbon$

$6.023\times10^{23}atoms are present in 12 of carbon$

$10^{23} atom will be present in=\frac{10^23}{6.023\times10^{23}}\times 12$ =

$1.992 g$

Hence b has the highest mass

The total mass of reactants and products in a reaction will always be the same. This law is known as:

This law is known as the law of conservation of mass. It states that in a closed system, the mass of the system cannot change over time i.e, the mass of the reactants must equal the mass of the products.

The atomic mass of carbon is:

The atomic mass of carbon is

$12.0107$

$u$ .

16 g of oxygen and 3 g of hydrogen are mixed and kept at 760 mm pressure and 0. The total volume occupied by the mixture will be nearly
(A) 22.4 litre
(B) 33.6 litres
(C) 448 litres
(D) 44800 ml

The correct option is D.

Given,

$16g$ of oxygen and

$3g$ of hydrogen

$pressure=760$

So, the mole of oxygen is

$\dfrac{1}{2}as\dfrac{16}{32}$ and the hydrogen is

$\dfrac{3}{2}$

Therefore the total moles are:

$\dfrac{1}{2}+\dfrac{3}{2}=2$

Now,

$PV=nRT$

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

$\dfrac{2\times0.0821\times273.15}{1}=44800ml$

What is hydrogen azide?

Hydrazoic acid is also known as hydrogen azide, or azoimide, chemical formula

$H{N_3}$ .

It is colourless, volatile & explosive liquid at room temperature and pressure.

It is a compound of nitrogen & hydrogen and it therefore a pnictogen hydride.

It was first isolated in

$1890$ by Theodor curtius.

What are the conditions for something to be called as matter?

Solution:-

The condition for something to be called matter is-

It should occupy some space and should have mass.

Define Avogadro's law. Taking a suitable example prove that it is not in contradiction with Dalton's Atomic Theory.

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The percentage of $3$ element are as follows in calcium carbonate $(CaCO_{3})$ Calcium $40\%$ , carbon $12\%$ , Oxygen $48\%$ . if law of constant proportion is true what weight of these element will be present in $1.5 g$ of another sample of calcium carbonates $(CaCO_{3})$

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An oxide $M_2O_3$ contains 10.3 % oxygen by weight. Calculate the atomic weight of M.

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A solution contains 50 g of sugar in 350 g of water. Calculate the concentration of solution in terms of mass by mass percent of the solution.

$\textbf{Step 1: Determine the mass of solution}$

The mass of the solution can be calculated as shown below.

$\textbf {Mass of solution = Mass of solute + Mass of solvent}$

Here, sugar is a solute and water is a solvent.

$\textbf{Mass of solution} = 50g+350g$

$= 400g$

Thus, the mass of the solution is 400 g.

$\textbf{Step 2: Determine concentration of the solution in terms of mass by mass percentage}$

The formula for the calculation of the concentration of the solution in terms of mass by mass percentage is given below.

$Mass\%=\dfrac{Mass \ of \ solute}{Mass \ of \ solution}\times 100$

$=\dfrac{50g}{400g}\times 100$

$=12.5\%$

Therefore, the concentration of the solution in terms of mass by mass percentage is

$12.5\%$

Chlorine occurs in nature in two isotope forms with masses 35u and 37u. The percentage of $^{35}\textrm{Cl}$ is 75%. Find the average atomic masses of chlorine atoms.

$\bullet$ Natural sample of chlorine contains 75% of

$Cl_{17}^{35}$ and 25% of

$Cl_{17}^{37}$ .

$\bullet$

$Average\ Atomic\ mass = \dfrac{\sum[(percentage\ of\ isotopes) \times (Mass\ of\ isotopes)]}{100}$

$\bullet$

$Thus, the\ average\ atomic\ mass\ of\ chlorine = \dfrac{(75 \times 25) + (25 \times 37)}{100} = 35.5$

$\textbf{Conclusion:}$

Therefore, the average atomic masses of chlorine atoms is 35.5.

Calculate the mass of 0.125 mol NaOH

$\textbf{Step 1: Determine the molecular mass of NaOH}:$

The molecular mass of NaOH can be calculated as,

$NaOH= (1\times Na)+(1\times O)+(1\times H)$

$=23+16+1=40$

Thus, the molecular mass of NaOH is 40.

$\textbf{Step 2: Determine the mass of NaOH}:$

The mass of NaOH can be calculated as shown below.

$Number \ of \ moles=\dfrac{Given \ mass}{Molecular \ mass}$

$0.125=\dfrac{Given \ mass}{40}$

$Given \ mass=0.125\times 40=5.0 \ g$

Therefore, the mass of

$NaOH$ is

$5.0 \ g$ .

What is one amu or one 'u'?

1 amu or 1 u

$= \cfrac{1}{12}$ th of the mass of an atom of carbon -12.

Define atomic mass unit.

An atomic mass unit is a unit of mass used to express atomic and molecular weights, equal to one twelfth of the mass of an atom of carbon-12.
OR

The atomic mass unit (amu) is defined as 1 / 12th of the mass of an atom of carbon.

$1 \,a.m.u. = 1.67 \times 10^{-24}g = 1.67 \times 10^{-27}kg$

Define the atomic mass unit.

An atomic mass unit is a unit of mass, equal to

$\frac {1}{12^{th}}$ the mass of an atom of the carbon-12 and used to express the mass of atomic and subatomic particles.

An atomic mass unit (symbolized AMU or amu) is defined precisely as

$\frac{1}{12}$ the mass of an atom of carbon -12.

Calculate the weight of ammonia gas required for reacting with sulphuric acid to give 78 g of fertilizer ammonium sulphate.

$\textbf{Step 1: Write the balanced chemical equation for the given reaction}$

The balanced chemical equation for the reaction of ammonia gas with sulphuric acid is as follows:

$2NH_3+H_2SO_4\rightarrow (NH_4)_2SO_4$

$\textbf{Step 2: Determine the molecular weight of the substances}$

The molecular weight of ammonia

$(NH_3)$ is

$14+3=17 \ g$

The molecular weight of ammonium sulphate

$[(NH_4)_2SO_4]$ is

$2\times(14+4)+32+4\times16=132 \ g$

$\textbf{Step 3: Determine the weight of ammonia gas}$

In the reaction,

$\mathop {2{\rm{N}}{{\rm{H}}_3}}\limits_{2 \times 17 = 34\;{\rm{g}}} + {{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \to \mathop {{{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}\limits_{{\rm{132}}\;{\rm{g}}}$

Here, the formation of 132 g of ammonium sulphate requires 34 g of ammonia

$(NH_3)$ .

Thus, the formation of 78 g of ammonium sulphate will require

$\dfrac{{34}}{{132}} \times 78\;{\rm{g}} = 20.09\;{\rm{g}}$ of ammonia

$(NH_3)$ .

$\textbf{Conclusion:}$

Therefore, 20.09 g of ammonia is required to produce 78 g of fertilizer ammonium sulphate.

$\text{The following questions are about one mole of sulphuric acid}\ [H_{2}SO_{4}]$ .
find the number of gram hydrogen atoms in 1 mole of $[H_{2}SO_{4}]$ .

$\textbf{Given data:}$

Number of moles of $H_2SO_4$ = 1 mole

we know that, 1 mole of $H_2SO_4$ contains 2 mole of hydrogen atoms.

$Number\ of\ mole\ of\ hydrogen = \dfrac{Mass\ of\ hydrogen}{Atomic\ mass\ of\ hydrogen}$

$2 = \dfrac{Mass\ of\ hydrogen}{1}$

$Mass \ of\ hydrogen = 2 g$

$\textbf{Conclusion:}$

Therefore, the number of gram hydrogen atoms in 1 mole of $H_2SO_4$ is 2 g.

During an experiment the students were asked to prepare a 10% solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water while Sarika prepared it by dissolving 10g of sugar in water to make 100g of the solution. Compare the mass % of the two solutions.

Solution made by Ramesh
Mass

$=\left ( \frac {10}{10+100} \right )100=\frac {10}{110}\times 100=9.09$ %
Solution made by Sarika
Mass

$=\frac {10}{100}\times 100=10$ %

Give reason: Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

The conversion of metal carbonates and sulphides to metals is very difficult and economically, it is worst. But metal oxides to metal conversion is easy. So they are converted to oxides.

The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale. If true enter 1, else enter 0

The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). It is defined as one-twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state and at rest. It has a value of

$1.66\times10^{-27} kg$ .

Mass and percentage by weight do not change with ________ (temperature/volume).

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The solubility of salt at $30^{\circ}C$ isCalculate the weight (g) of water required to prepare a saturated solution containing 90 g of salt.

Solubility (S)

$=$ 50
Amount of salt (solute)

$= w_1 = 90 g$
Amount of water (solvent)

$=w_2=?$
Solubility (S)

$=w_1/w_2 \times 100$ ............. (4)
Weight of water (solvent)

$=w_2 = 100 w_1/S$

$=100 \times 90/50 = 180 g$

If 11 g of oxalic acid is dissolved in 500 mL of solution (density $=$ 1.1 g mL $^{-1}$ ), what is the mass % of oxalic acid in solution?

11 g of oxalic acid is present in 500 mL of solution.

Density of solution

$=1.1 g mL^{-1}$

Mass of solution

$=(500 mL) \times (1.1 g mL^{-1})$

$=550 g$

Mass of oxalic acid

$=$ 11 g

Mass % of oxalic acid

$\displaystyle = \frac{11}{550} \times 100 =2$ %

$\displaystyle \begin{matrix}Isotope &per\ Natural\ abundance &Molar\ mass \\^{1}H &99.985 &1 \\ ^{2}H &0.015 & 2\end{matrix}$

The average atomic mass of hydrogen using the following data is _________.

Data:-

Isotope	Natural abundance	Molar mass
$^1H$	99.985	1
$^2H$	0.015	2

Formula:-

Average atomic mass=

$\cfrac {Atomic\quad mass\quad of\quad ^1H\times abundance+Atomic\quad mass\ of\quad ^2H\times abundance}{100}$

$=\cfrac {1\times 99.985+2\times 0.015}{100}$

$=\cfrac {99.985+0.030}{100}$

$=1.00015$ .

The atomic mass of $_{11}^{23}\textrm{Na}$ in amu is :

$_{11}^{23}\textrm{Na}$ has the mass number 23, so its atomic mass is 23 amu.

The percent composition by mass of the phosphorus in calcium phosphate $Ca_3(PO_4)_2$ is $x$ . Then, the value of $x$ is :

The molar mass of

$Ca_3(PO_4)_2 = 310$ g
%

$Ca$

$=120\times \dfrac {100}{310}=38.71$
%

$P$

$=2\times 31 \times \dfrac {100}{310}=20$
%

$O$

$=8\times 16 \times \dfrac {100}{310}=41.29$
Thus,

$x$ is

$20$ .

Boron has two stable isotopes, $^{10}$ B(19%) and $^{11}$ B(81%). The atomic mass that should appear for boron in the periodic table is ______ .
(Enter the nearest integer value)

The aluminum sulphate hydrate $[Al_2(SO_4)_3\cdot xH_2O]$ contains $8.20$ percent $Al$ by mass. The value of $x$ , that is, the number of water molecules associated with each $Al_2(SO_4)_3$ units is ________. (write the ans in the form of x/3)

Molar mass of

$Al_2(SO_4)_3\cdot xH_2O=342+18x$ .

$\%$ of

$Al=\displaystyle\frac{54\times 100}{342+18x}=8.2$ .

$x=18$ .

Which of the following pairs have equal mass?
(1) $1$ amu of $^{12}\textrm{C}$ and $1$ amu of $^{13}\textrm{C}$
(2) $1$ mol of $_{11}^{23}\textrm{Na}$ and $1$ mol of $_{11}^{24}\textrm{Na}$
(3) $0.5$ mol of $_{2}^{4}\textrm{He}$ and $1$ mol of $_{1}^{2}\textrm{H}$
Answer must be the sequence number of the pair.

In element representation

$_{ Z }^{ M }{ E }$ ,

$M$ is the mass. Therefore out of the given options, 0.5 mol of

$_{ 2 }^{ 4 }{ He }$ has mass

$0.5\times4=2$ and

$_{ 1 }^{ 2 }{ H }$ has mass 2, both are equal. Therefore option (3) is correct.

$Al_2(SO_4)_3.xH_2O$ has $8.20$ % $Al$ by mass. The value of $x/3$ is _________.

According to given conditions for

$Al_2(SO_4)_3. xH_2O$ ,

$\dfrac{54\times100}{(18x+3\times96+54)} =8.20$

$\therefore x\approx 18$ .

$\therefore \dfrac{x}{3}=\dfrac{18}{3} =6$

Atomic mass unit is also called Dalton(Da). Explain.

The unified atomic mass unit or dalton (Da) is the standard unit that is used for indicating mass on an atomic or molecular scale (atomic mass). One unified atomic mass unit is approximately the mass of one nucleon (either a single proton or neutron). It is defined as one-twelfth of the mass of an unbound neutral atom of carbon-12 in its electronic ground state.

Analysis of a metal chloride $XCl_3$ shows that it contains $67.2\%$ $Cl$ by mass. The molar mass of $X$ is __________.

Let, molar mass of X is

$M$ .
So,

Mass % of

$Cl=\dfrac{3\times35.5\times100}{(M+3\times35.5)} = 67.1$

$M = 52 g$ .

Thus, the molar mass of X is

$52$ g

Disilane, $Si_2H_x$ , is analyzed and found to contain $90.28\%$ of silicon by weight, then the value for $x$ is $(Si=28)$ :

According to the given conditions,

$90.28\%$ of silicon by weight:

So, Weight of silicon

$/$ Total molecular weight

$= 0.9028$

$\dfrac{28\times2}{56+x} = 0.9028;$

$\implies x=6$

$Na_2SO_3\cdot yH_2O$ has $50\%$ $H_2O$ by mass. Thus, $y$ is :

$\underbrace{ Na_2SO_3}\cdot \underbrace{yH_2O}$

$\downarrow$

$126g$

$18y$

$\displaystyle\frac{18y}{126+18y}=\displaystyle\frac{50}{100}$

$\therefore y=5$

Match $\%$ of carbon in List-I with the compound given in List-II.

20%

$= CN_2OH_4 = 12\times 100/60$
75 %

$= CH_4 = 12\times 100/16$
12 %

$=CaCO_3 = 12 \times100/100$
27.3%

$=CO_2 = 12 \times 100/44$
52.2%

$=C_2H_6O$

When $2.495$ $g$ of $CuSO_4\cdot xH_2O$ (molar mass $249.5$ ) is heated, $0.05$ mole of $H_2O$ is lost. Thus, $x$ is:

According to given conditions,
Moles of compounds

$=\dfrac{mass}{molar\ mass}=\displaystyle\dfrac{2.495}{249.5}=0.01$ mol

$0.01x$ of

$H_2O$

$=0.05$ mol (given)

$\therefore x=5$

A solution _______ $\%$ by mass of solvent is $16.67\%$ by mass of solution. Write the value in the form of $x\times10$ , where $x$ is:

Let,

Mass of solvent

$= 100$ g

Mass of solute

$= x$ g

The percentage by mass of solution will be

$\dfrac {\text{mass of solute}} {\text{mass of solution}} \times 100= \dfrac {x} {100+x} \times 100 = 16.67$ %.

$\Rightarrow x =20$ g

Percentage mass by solvent

$=\dfrac {\text{mass of solute}} {\text{mass of solution}} \times 100=\dfrac{20}{100}\times 100 = 20$ %

Value of

$x =$ 2.

Total number of ions in aluminium phosphate is :

$AlPO_4\rightleftharpoons Al^{3+}+PO^{3-}_4$
Total 2 ions are there.

There are_______moles of anion in one mole of sodium sulphate.

$Na_2SO_4 \rightarrow 2Na^+ + SO_4^{-2}$

If at a particular temperature, the density of $18\ M \ H_2SO_4$ is $1.8$ g cm $^{-3}$ . Percentage concentration (%) by mass of solution is :

$Molarity=\dfrac { { w }_{ solute } }{ { M }_{ solute }\times V(litre) }$

Let the volume of solution be 1 L.

$18=\dfrac { { w }_{ solute } }{ 98\times 1 }$

${ w }_{ solute }=18\times 98gm$

% concentration by mass=

$\dfrac { w_{ solute } }{ { w }_{ solution } } \times 100$

Density of solution is 1.8g/

${ cm }^{ 3 }$ or 1800g/L.

The volume of solution is 1 L.

So, weight of solution is 1800g.

% composition by mass

$= \dfrac{18 \times 98}{1800} \times 100 = 98$ %

Matter is neither .......... nor .......... in a chemical reaction.

Matter is neither

$\underline {created}$ nor

$\underline {destroyed}$ in a chemical reaction. This is the law of conservation of mass.

Calculate the mass percent of $O$ present in sodium sulphate $\displaystyle \left( { Na }_{ 2 }{ SO }_{ 4 } \right)$ .

The molar mass of sodium sulphate is

$\displaystyle 2(23)+32+4(16) = 46+32+64 = 142 \: g/mol$

Mass percentage

$= \dfrac{\text{Mass of oxygen}}{\text{Mass of }Na_2SO_4}\times 100$

The mass percentage of oxygen in sodium sulphate

$\displaystyle = \frac {64}{142} \times 100 = 45.07\%$ .

In diammonium phosphate, $(NH_4)_2HPO_4$ , the percentage of $P_2O_5$ is

Two mole of

$(NH_4)_2HPO_4$ gives one mole of

$P_2O_5$ so
% of

$P_2O_5 = 142\times100/264 = 53.78$ %.
The nearest integer value is 54.

Describe law of conservation of matter.

The law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as system mass cannot change quantity if it is not added or removed. Hence, the quantity of mass is "conserved" over time. The law implies that mass can neither be created nor destroyed, although it may be rearranged in space, or the entities associated with it may be changed in form, as for example when light or physical work is transformed into particles that contribute the same mass to the system as the light or work had contributed.

The mass of 94.5 ml of a gas at S.T.P. is found to be 0.2231 g. Calculate its molecular mass.

Given:-

Mass of gas=

$0.2231g$

Volume of gas at S.T.P=

$94.5ml=94.5\times10^{-3}$ litre or

$dm^3$

To find:- Molecular mass

Solution:-

$\cfrac {Volume\quad occupied\quad by\quad gas\quad at\quad STP\quad in\quad dm^3}{22.4dm^3}$ =Number of moles

$\Rightarrow \cfrac {94.5\times 10^{-3}dm^3}{22.4 dm^3}$ = Number of moles

$\Rightarrow$ Number of moles=

$0.0042$

Number of moles=

$\cfrac {Weight}{Molecular\quad mass}$

$\therefore$ Molecular mass=

$\cfrac {0.2231g}{Number\quad of\quad moles}=\cfrac {0.2231g}{0.0042mole}$

$=53.11g/mole$

$\therefore$ Molecular mass=

$53.11g/mol$ .

16 grams of $NaOH$ is dissolved in 100 grams of water at 25 $^o$ C to form a saturated solution. Find the solubility and weight percentage of the solution.

$16 \,g NaoH, 100 g$ water

weight percent of

$NaoH = \dfrac{16}{100+16}\times 100=13.79$ %

weight percent of water

$= \dfrac{100}{100+16}\times 100=86.21$ %

Solubility =

$\dfrac{mass\, of\, compound }{mass\, of\, solvent}$

$=\dfrac{16}{100}$

$= 16 g / 100g$

Compounds are formed by chemically combining elements in a .................... proportion by weight.

Compounds are made up of two or more elements combined in definite quantities and in definite proportions by chemical bonds. The law of definite proportions states that a given chemical compound always contains its component elements in fixed ratio by mass or weight and does not depend on its source and method of preparation.

Why do atomic masses of most of the elements in atomic mass units involve fraction?

Atomic mass unit is defined as

$\cfrac {1}{12^{th}}$ of mass of an atom of carbon-12.

Many naturally occurring elements exist in the form of one or more isotopes. Isotopes are the atoms of the same element having the same atomic number but different mass numbers.

While taking the weighted average of all naturally occurring isotopes, the result may be a fractional number.

For example, carbon occurs naturally in three isotopic forms

$C-12, C-13, C-14$ . Its average atomic mass is

$12.01 \ u$ .

If $10 l$ of oxygen contains x molecules, then $10 l$ of ozone contains ______ molecules at the same temperature and pressure.

According to Avogadro's law equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

If the volume for both oxygen and ozone = 10 liters, then their number of molecules also the same.

So, the correct answer is x

Calculate the mass of urea $NH_2CONH_2$ required in making 2.5 kg of 0.25 molal aqueous solution.

0.25 molal aqueous solution of urea means that 0.25 mole of urea are present in 1 kg of water

Moles of urea =

$0.25\, mole$

Mass of solvent (water) = 1 kg = 1000 g

Molar mass of urea (

$NH_2CONH_2$ ) = 60 g mol

$0.25 mole of urea = 0.25\, mol\, \times 60 g\, mol\, 1 = 15 g$

${\text Total\, mass\, of\, solution = 1000 + 15 g = 1015 g = 1.015 kg}$

Thus; 1.015 kg of solution contain urea =

$15 g \times 2.5 kg$ of solution will require urea

$\dfrac{15g}{1.015 kg } \times 2.5 kg$ = 37 g

Fill in the blanks with suitable words:
The atomic mass of an atom is $23$ and its atomic number is $11$ . The atom has ________ neutrons.

The atomic mass is the sum of the number of protons and neutrons. The atomic number is the number of protons.

Number of neutrons

$=$ Atomic mass

$-$ Atomic number

$=23-11=12$

Fill in the blanks with suitable words.
Average atomic mass of chlorine is ____________.

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In the brown ring complex, the oxidation state of iron is:

Nitrates give brown fumes with concentrated $H_2SO_4$ due to formation of $NO_2$ . This is intensified upon adding copper turnings. Nitrate ion is confirmed by adding an aqueous solution of the salt to $FeSO_4$ and pouring concentrated $H_2SO_4$ slowly along the sides of the test tube, which produces a brown ring around the walls of the tube, at the junction of the two liquids caused by the formation of ${[Fe{(H_2O)}_5(NO)]}^{2+}$ .

State Avogadro's law?

"Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules".

It shows us the correlation between the number of molecules and volumes of gases.

a) State any three findings of modem atomic theory.
b) Write down any two applications of Avogadro's law.

a) The findings of modern atomic theory are:

(i) Atom is considered to be divisible particle.
(ii) Atoms of the same element may not be similar in all respects
eg: Isotopes

$Cl _{17}^{35}$ and

$Cl _{17}^{37}$
(iii) Atoms of different elements may be similar in some respects.
eg: Isobars

$Cl _{40}^{18}$ and

$Cl _{40}^{20}$
(iv) Atom is the smallest particle which takes part in chemical reactions.

b) Applications of avagadro's law are:

Determines the atomicity of the gas.
Determines the molecular formula of the gas.
Explains Gay lussac's law.

Find the concentration of solution in terms of weight percent if $20$ grams of common salt is dissolved in $50$ grams of water.

Percentage composition =

$\dfrac{weight\: of \: the\: common\: salt}{weight\, of\, common\, salt\, +\, weight\,of\,water} \times 100$

$\dfrac{20}{20+50} \times 100$

$28.6$ %

Details of some gases taken at ${273}\ K$ temperature and $1$ atm pressure are given in the table.
Gas Volume (L) Number of moles
Hydrogen 224 10
Helium 112 5
Oxygen 224 10
Ammonia 56 2.5
(a) On the basis of which gas law, the volume and the number of moles are given in the table?
(b) Represent this gas law in mathematical form.
(c) Among these, if the pressure of hydrogen gas is changed to $2$ atm. Calculate the new volume.

(a) Avogadro's law
(b)

$V\propto n(PT$ constant)
(c)

${V}_{1}=224L,\ {P}_{1}=1atm,\ {P}_{2}=2atm,\ {V}_{2}=$ ?

${P}_{1}{V}_{1}={P}_{2}{V}_{2}$ or

${V}_{2}={P}_{1}{V}_{1}/{P}_{2}$

$=\dfrac{1\ atm\times 224\ L}{2\ atm}=112\ L$

Find the concentration of solution in terms of weight percent if $20g$ of common salt is dissolved in $50g$ of water.

Solute

$=20g$

Solvent

$=50g$

Solution

$=20+50=70g$

Concentration of solution

$=\cfrac{20}{70}\times 100\\=28.57\%$

Calculate the weight of lime ( $CaO$ ) that can be obtained by heating 200 kg of limestone which is 93% pure. How many moles of impure potassium chlorate of 75% purity is required to produce 48 g of oxygen?

$CaCO_3 \rightarrow CaO + CO_2$

moles of

$CaCO_3$ in 200 kg = 1000 moles

moles of CaO produced = 1000 moles

weight of CaO produced = 1000* 56 = 56000g

limestone is 90% pure

therefore the amount of CaO produced =

$56000* \dfrac{93}{100}$ = 52080g or 52.08 kg.

we have,

$KClO_3 \rightarrow KCl + 3/2 O_2$

moles of

$O_2$ in 48 g =

$\dfrac{48}{32}$ = 1.5

that means 1 mole of

$KClO_3$ is dissociating.

1 mole of

$KClO_3$ = 122.5 g

so, amount of impure

$KClO_3$ =

$122.5*\dfrac{100}{75}$ = 163.3 g

The relative density of a mixture of nitrogen and oxygen is 14.4 (H = 1) and the relative densities of nitrogen and oxygen are 14.0 and 16.0 (H =1)respectively. Calculate the composition of the mixture (i) by volume and (ii) by mass.

(i) Let 1 mL of the mixture contain x mL of

$N_2$ and (1 - x) mL of

$O_2$ .
Mass of x mL of

$N_2 =140 \times x =14 x g$ (Mass

$= d \times V$ )
Mass of (1 - x) mL of

$O_2 =16.0 \times (1- x)= (16.0 -16 x) g$
Total mass of mixture

$=14 x +16.0 -16 x$
So,

$14x + 16.0 - 16x=14.4 \times 1=14.4$
or x = 0.8 , i.e., 80% by volume
Oxygen

$=1- x = (1 - 0.8) = 0.2$ , i.e., 20% by volume.
(ii) Let 1 g of the mixture contain x g of

$N_2$ and (1- x) g of oxygen.
Volume of x g of

$N_2 = \frac{x}{14.0} \left ( V=\frac{mass}{density} \right )$
Volume of (1-x) g of

$O_2=\frac{(1-x)}{16.0}$
Total volume of the mixture

$=\frac{1}{14.0}+\frac{(1-x)}{16.0}$
So,

$=\frac{1}{14.0}+\frac{(1-x)}{16.0}=\frac{1}{14.4}$
or x=0.7778, i.e., 77.78% by mass
Oxygen

$=(1-x)=(1-0.7778)= 0.2222$
i.e.,

$22.22\%$ by mass.

2 g mixture of glucose and sucrose is dissolved in 1 litre water at 298 K to develop an osmotic pressure of 0.207 atm. Calculate percentage composition of glucose and sucrose by mole as well as by mass.

Let (x)g be glucose and (2-x) g be sucrose

$\Pi$ =CRT

0.207=

$(\dfrac{x}{180\times1}+\dfrac{2-x}{342\times1}$ )

$\times$ 298

$\dfrac{x}{180}+\dfrac{2-x}{342}$ =0.0084

Solving for x , we get x=0.97g

weight of glucose= 0.97 g , weight of sucrose = 1.03g

moles of glucose =

$\dfrac{0.97}{180}$ =5.38

$\times 10^{-3}$

moles of sucrose=

$\dfrac{1.03}{342}$ = 3.01

$\times 10^{-3}$

Find the amount of zinc in gram necessary to produce $224ml$ of ${H}_{2}$ by the reaction with sulphuric acid.

Given, Reaction between

$Zn$ and Acid to produce

$H_2$ is

$Zn+H_2SO_4\longrightarrow ZnSO_4+H_2$

From above reaction, it is clear that

$1$ mole of

$Zn$ produces

$1$ moles of

$H_2$

$\therefore$

$224$ ml of

$Zn$ was required to produce

$224$ ml

$H_2$ .

Now we know

$1$ mole=

$22.4L$ of

$Zn$

$\Rightarrow 224ml$ of

$Zn=224\times 10^{-3}L$ of

$Zn$

$\Rightarrow 224\times 10^{-3}L$ of

$Zn=\cfrac {224\times 10^{-3}}{22.4}$ moles of

$Zn$

$\therefore$ No. of moles required=

$10^{-2}$ moles of

$Zn$

We know,

$1$ mole of

$Zn=65 gms$ of

$Zn$

$\Rightarrow 10^{-2}$ moles of

$Zn=65\times 10^{-2}gms$ of

$Zn$

$\therefore$ Amount of Zinc in gms necessary to produce

$224ml$ of

$H_2=0.65gms$ .

A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen.Calculate the percentage composition of the compound by weight.

% composition of oxygen in compound

$=\cfrac { wt\quad of\quad oxygen }{ wt\quad of\quad compom } \times 100$

$\therefore \quad 0.24gm$ of compound contains

$0.096gm$ of oxygen

$\because \quad 1gm$ of compound contains

$\cfrac {0.096}{0.24} gm$ of oxygen

$\because \quad 100gm$ of compound

$\cfrac {0.096}{0.24} \times 100\ gm$ of oxygen

$=40$ %

$\because$ % composition of Boron in compound

$=(100-40)$

$=60$ %

Calculate the weight of lime ( $CaO$ ) obtained by heating $200kg$ of 95% pure lime stone ( $Ca{CO}_{3}$ ).

$Ca{CO}_{3}\rightarrow CaO+{CO}_{2}$

∵ 100 kg impure sample has pure

$CaCO_3$ = 95 kg

∴ 200 kg impure sample has pure

$CaCO_3$ = (95 x 200 / 100) = 190 kg.

$CaCO_3 \rightarrow CaO + CO_2$

∵ 100 kg

$CaCO_3$ gives

$CaO$ = 56 kg.

∴ 190 kg

$CaCO_3$ gives

$CaO$ = (56 x 190 / 100) = 106.4 kg.

The density of iron crystal is $8.54$ gram $cm^{-3}$ . If the edge length of unit cell is $2.8$ $A^o$ and atomic mass is $56$ gram $mol^{-1}$ , find the number of atoms in the unit cell. (Given: Avogadro's number $=6.022\times 10^{23}$ , $|A^o=1\times 10^{-8}$ cm).

The edge length

$\displaystyle a=2.8 \ A^o \times (1 \times 10^{-8} \ cm/A^o) = 2.8 \times 10^{-8} \ cm$

Density

$\displaystyle \rho = \dfrac { Z M}{N_0a^3 }$

$\displaystyle 8.54 \ g/cm^3 = \dfrac {Z \times 56 \ g/mol}{6.022 \times 10^{23} \ /mol \times ( 2.8 \times 10^{-8} \ cm)^3}$

$\displaystyle Z=2$
The number of atoms in the unit cell is

$\displaystyle 2$ .

At 50C, liquid ${ NH }_{ 3 }$ has ionic product is ${ 10 }^{ -30 }$ . How many amide $\left( N{ H }_{ { 2 }^{ - } } \right)$ ions are present per ${ mm }^{ 3 }$ in pure liquid ${ NH }_{ 3 }$ ? (take ${ N }_{ A } = 6 \times \ { 10 }^{ 23 }$ )

Ionic product of

$NH_3,\ K_{NH_3}=[NH_4^+][NH_2^-]=10^{-30}$

$[NH_2^-]=\sqrt{10^{-30}}=10^{-15}\ M$

Since,

1 litre contains

$10^{-15}\ moles$

then

$10^6\ mm^3$ contains

$10^{-15}\ moles$

Therefore

$1\ mm^3$ contains

$=\frac{10^{-15}}{10^6}=10^{-21}\ moles$

Number of

$NH^{2-}\ ions=6\times10^{23}\times10^{-21}=600\ ions$

The self ionisation constant for $HCOOH$ is ${10}%{-6}$ what percentage of $HCOOH$ are converted to $HCOO$ ion. The density of $HCOOH$ is $1.22g$ ${mL}^{-1}$ .

R.E.F image

$HCOOH \rightleftharpoons HCOO^{\circleddash }+4^{\oplus }$

$[HCOOH] = \dfrac{n}{v} = \dfrac{mass}{mosot}\times \dfrac{1}{v}$

$\dfrac{mass}{volume} = density$

$= \dfrac{1.22\times 10^{3}}{46}$

$= 26.52\,M$

$HCOOH\rightleftharpoons HCOO^{-}+H^{+}$

$\dfrac{x^{2}}{(1-x)} = k = 10$

$x$ is small & it can be neglect

$x = \sqrt{10}$

$\%$ dissociation

$= \dfrac{[HCOO^{-}]}{[HCOOH]}\times 100 = \dfrac{\sqrt{10}}{26.52}\times 100 = 11.92\%$ Ans

1120526_869898_ans_93d462a382434e07b8e46ddb7334b81f.jpg

Lithium exists in nature in the form of two isotopes, Li-6 and Li-7 with atomic masses 6.0151u and 7.0160u and the percentages 8.24 and 91.76 respectively. Calculate average atomic mass.

Atomic masses of

${ Li }^{ 6 }$ and

${ Li }^{ 7 }$ are

$6.0151u$ and

$7.0160u$ respectively.

$\therefore$ Average atomic mass

$=\dfrac { 6.0151\times 8.24+7.0160\times 91.76 }{ 100 } =6.9335u$

What should be the number of ${{\text{H}}^{\text{ + }}}$ ions in 1 mL of distilled water, if the number of ${{\text{H}}^{\text{ + }}}$ ions in L is ${\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{16}}$ ?

$H_2O\rightarrow H^++OH^-$

As one water molecule gives one

$H^+$

$1L$ no. of

$H^+$ ions

$=6.023\times10^{16}$

So, In

$1ml$ it will be

$=\cfrac{6.023\times 10^{16}}{1000}\\n=6.023\times10^{13}$

State the law of conservation of mass. State the main points of Landolt's experiment for experimental evidence of the law.

Refer to image.

Law of conservation of mass states that in any chemical reaction, the initial weight of reactants is equal to the final weight of the products. In other words, Mass can neither be created nor destroyed during a chemical reaction but it only changes from one form into another.

Main points of Landolt's Experiment were:-

(1) A specially designed H-shaped tube is taken with its one limb having Sodium chloride solution and the other have

$AgNO_3$ . This, in the first place was weighed after sealing.

(2) The reaction was made to take place and then the tube was weighed.

(3) The mass of the tube in both the case was found to be the same. Thus redefining the Law of Conservation of Mass.

1099060_1004675_ans_64cc8ac6559a4ae4b83b0c254290cb2c.PNG

Discuss about determining mass percentage of oil from oil seed.

REQUIREMENTS:- Steam generator (Copper Vessel), round bottom flask

$(500 ml)$ , conical flask, condenser, glass tubes, iron stand, sand bath, separatory funnel, tripod stands, burners, Ajwain(Carum), Petroleum ether

$(60-80°C)$ ,Saunf(Aniseed).

PROCEDURE:-

Set the apparatus as shown in the picture of Experimental Setup.
The apparatus consists of a steam generator connected to the round bottom flask through a glass inlet tube. The flask is connected to a water condenser through a glass outlet tube. Condenser is further attached to a receiver through an adaptor.
Take about $750 ml$ of water in the steam generator and start heating to produce steam. In the round bottom flask take about $75 gm$ of crushed saunf.
A vigorous current of steam from steam generator is passed through the round bottom flask. A part of the steam condenses in the round bottom flask. As more and more steam is passed, the steam volatile components of saunf pass through the condenser along with steam.
These contents on condensation are collected in the receiver. The contents in the round bottom flask may be heated by a bunsen burner to prevent excessive condensation of steam.
The process of steam distillation is continued for about half an hour.
Transfer the distillate to a separating funnel and extract with $20 ml$ portions of petroleum ether $3$ times. Combine the petroleum ether extracts in a $250 ml$ conical flask and dry it with the help of anhydrous sodium sulphate.
Remove the solvent from the dried filtrate by careful distillation in a water bath. The essential oil is left behind in the distillation flask. Find the weight of the extracted essential oil. Note the colour, odour and weight of the essential oil.

Calculate the percentage loss of mass of hydrated copper[II] sulphate $[CuSO_4.5H_2O]$ when it is completely dehydrated.
$CuSO_4.5H_2O \to CuSO_4+5H_2O$
[Atomic weights are $Cu = 64, S=32, O=16, H=1$ ]

The molecular weight of water

$\displaystyle H_2O = 2 (1)+16=18$ .

When 1 mole of

$\displaystyle CuSO_4 \cdot 5 H_2O$ is heated, 5 moles of water are lost.

The molecular weight of

$\displaystyle CuSO_4 \cdot 5 H_2O=64+32+4(16)+5(18)=250$ g/mol

Mass of water lost

$\displaystyle = 5 (18)=90$ g

Percentage loss of mass of

$\displaystyle CuSO_4 \cdot 5 H_2O = \dfrac {90}{250} \times 100=36$ %.

92 g mixture of $CaCO_3$ and $MgCO_3$ heated strongly in an open vessel. After complete decomposition of the carbonates it was found that the weight of residue left behind is 48 g. Find the mass of $MgCO_3$ in grams in the mixture.

$CaCO_3(s)+MgCO_3(s)\xrightarrow [ ]{\Delta } CaO(s)+MgO(s)+2CO_2(g)$

Thus on Reaction of

$1$ mole of

$CaCO_3$ &

$1$ mole of

$MgCO_3$ , the residue left is

$14$ mole of

$CaO$ &

$1$ mole of

$MgO$

Molecular wt of

$CaCO_3=100\ g$

$MW$ of

$MgCO_3=84\ g$

$M$ wt of

$CaO=56\ g\quad MW$ of

$MgO=40\ g$

Thus when a mixture of

$184\ g$ is heated, residue weight

$96\ g$

Also, when a mixture of

$92\ g$ is heated, residue weight

$48\ g$

A mixture of contains

$84\ g\ MgCO_3$

$\therefore \$ A mixture of

$92\ g$ contains

$42\ g\ MgCO_3$

Calculate the mass per cent of different elements present in sodium sulphate $(Na_2SO_4)$ .

Molar mass of

$Na_2SO_4$ =

$23 \times 2 + 32 + 4 \times 16= 142$

$Na = \dfrac{23 \times 2}{142} \times 100 = 32.4\%$

$S = \dfrac{32}{142} \times 100 = 22.5\%$

$O= \dfrac{64}{142} \times 100 = 45\%$

Claculate the mass % of different element in $Na_2SO_4$ ?

Molar mass of sodium sulphate

$Na_2SO_4 = [(2\times23) + (32) + 4(16)] = 142gm$

Mass % of an element = Mass of that element in the compound / Molar mass of the compound x 100

Mass % of sodium =

$\dfrac{46}{142}\times100 = 32.39%$

Mass % of Sulphur =

$\dfrac{32}{142}\times100 = 22.53$

Mass % of oxygen =

$\dfrac{64}{142}\times100 = 45.07$

In a Carius determination, $0.234g$ of an organic substance gave $0.334g$ of barium sulphate. Calculate the percentage of sulphur in the given compound. $[Ba = 137, S=32, O=16]$

Mass of Organic Substance = 0.234 g

Mass of Barium Sulphate = 0.334 g

Molecular Mass of

$BaS{O}_{4}$ = 233

$g {mol}^{-1}$

233 g of

$BaS{O}_{4}$ contain 32 g of Sulphur

0.334 g of barium sulphate would contain

$\dfrac{32}{233} \times 0.334$ = 0.0458 g

% of Sulphur =

$\dfrac{0.0458}{0.234} \times 100$ =

$19.6$ %

Calculate the no. of atoms in:
a) 0.5 mol atom of nitrogen
b) 0.2 mol molecules of hydrogen
c)3.2 g of sulphur (Z for sulphur=32)

$(a)$

$0.5$ mol of atom of nitrogen have

$= 6.022 \times 10^{23} \times 0.5 = 3.011 \times 10^{23}$

$(b)$

$0.2$ mol of molecules of hydrogen have

$= 6.022 \times 10^{23} \times 0.2 = 11.204 \times 10^{23}$

$(c)$

$3.2$

$g$ of

$0.1$ mol of sulphur have

$= 6.022 \times 10^{23} \times 0.1 = 6.022 \times 10^{22}$

State and explain law of conservation of mass.

Law of conservation of mass implies that mass can neither be created nor destroyed, although it may be rearranged in space, or the entities associated with it may be changed in form. For example, in chemical reactions, the mass of the chemical components before the reaction is equal to the mass of the components after the reaction.

A solution contains 40 grams of common salt dissolved in 345 grams the solution. Calculate the concentration percentage of the solution by mass percentage.

Mass (solute) = 40 grams

Mass ( solvent ) = 345 grams

Mass( solution) = Mass(solute) + Mass(solvent)

= 40 grams + 345 grams

= 385 grams

Therefore Mass % (solution) = [Mass(solute) / Mass(solvent)]

$\times 100$

$\frac{40}{385}$

$\times 100$

=11. 59%

0.246 g of an organic compound on complete combustion gave 0.198 g of carbondioxide and 0.1014 g of water, then the percentage composition of carbon and hydrogen in the compound respectively
1) 4.58, 21.95 2) 21.95, 4.58
3) 45.8, 2.195 4) 2.195, 45.8

The correct options is B

in the given question

mass of organic compound=0.246 g

mass of carbon dioxide= 0.198 g

mass of water= 0.1014 g

mass of water=0.1014 g

% of carbon:

$\frac{12}{44}\times \frac{0.198}{0.246}\times 100$

=21.95%

% of Hydrogen

$\frac{2}{18}\times \frac{mass of H_2O}{mass of compound}\times 100$

$\frac{2}{18}\times \frac{0.1014}{0.246}\times 100$

=4.58%

State and explain laws of chemical combination.

Law of Chemical Combination

$(1)$ Law of Conservation of Mass

In a chemical change, the total mass remains conserved. i.e. mass before the reaction is equal to mass after the reaction.

${ H }_{ 2 }(g)+{ Cl }_{ 2 }(g)\longrightarrow 2HCl$

Mass before reaction=

$2+71=73g$

Mass after reaction=

$2$ x

$36.5=73g$

$(2)$ Law of constant or Definite Proportions

All Chemical compounds are found to have constant composition irrespective of their method of preparation or source.

For g. in water

$({ H }_{ 2 }O)$ , hydrogen and oxygen combine in

$2:1$ molar ratio, this ratio remains constant whether it is tap water or river water.

$(3)$ Law of Multiple Proportions

When one element combines with the other elements to form two or more different compounds, the mass of one element which combines with a constant mass of the other, bear a simple ratio to one another.

For eg.

$C$ and

${ O }_{ 2 }$ reacts together to form

$CO$ and

${ CO }_{ 2 }$ . Here, fixed mass of carbon

$(12g)$ reacting with different masses of oxygen

$(16g$ and

$32g)$ . Therefore, ratio of the masses of oxygen has ratio

$16:32$ or

$1:2$

If atomic eq. 40% of $PCl_{ 5 }$ dissociated. Calculate equivalent molar mass.

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

Lct ni(initial moles of

$PCl_5)=1$

$1$

$0$

neq

$1-0.4=0.6$

$0.4$

$0,.4$

$\therefore$ molar mass =

$\sum$ mols tradition of complement

$\times$ molecules mass

$=\dfrac{0.4}{1.4}\times 71+\dfrac{0.4}{1.4}\times 137.5+\dfrac{0.6}{1.4}\times 208.5$

$=148.93$

Calculate the mass percentage composition of the elements in nitric acid : $(H=1;N=14,O=16)$

Molecular formula of Nitric acid=

$HNO_3$

Mass % of

$H$ =

$\cfrac {Mass\ of\ H\ in\ 1\ mole\ of\ HNO_3}{Molecular\ mass\ of\ HNO_3}\times 100$

$\cfrac {1}{1+14+(16\times 3)}\times 100$

Mass % of

$H$ =

$1.59$ %

Mass % of

$N$ =

$\cfrac {14}{1+14+(16\times 3)}\times 100$

$22.22$ %

Mass % of

$O$ =

$\cfrac {16\times 3}{1+14+(16\times 3)}\times 100$

$76.19$ %

Mass % of composition is

$H=1.59$ %

$N=22.22$ %

$O=76.19$ %

What is the law of conservation of mass?

It states that during a chemical reaction, mass is neither created nor destroyed. Thus, the total mass of the substances before the reaction and the total mass of the substances after the reaction remain the same.

Thus, no mass is lost during the chemical reaction, and the mass is conserved.

Also, for balanced chemical reaction, the total mass of the reactant is equal to the total mass of the product.

Calculate the total number of electrons present in $18\ mL$ of water.

Volume of water

$=18$

$mL$

Since denisty of water = 1 g/mL

$\therefore$ Mass of water

$=18$

$grams$

No. of moles of water

$=\cfrac{18}{18}=1$

$1$ molecule of

$H_2O$ has

$10$ electrons

$\therefore 1$ mole

$=6.023\times 10^{23}$ molecules

$\therefore$ No. of electrons

$=6.023\times 10^{23}\times 10=6.023\times 10^{24}$

Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.

Atomic mass is the sum of electron and neutron in the atoms. Atomic number gives the number of electrons So, to determine the atomic number the reference of atom is needed.

If we do not take the reference atom for atomic mass, then the mass values will be more complex to memorize.

The carbon atom is one of the reference atom. The relative mass a carbon atom was accepted as 12.

$1.20g$ sample of ${Na}_{2}{CO}_{3}$ and ${K}_{2}{CO}_{3}$ was dissolved in water to form $100ml$ of a solution. $20ml$ of this solution required $40ml$ of $0.1N$ $HCl$ for complete neutralization. Calculate the weight of ${Na}_{2}{CO}_{3}$ in the mixture.

Total weight of the mixture(Na2CO3+K2CO3)=1.20g

Volume of the solution=100ml

Volume of solution taken for neutralisation V2= 20 ml

Volume of HCl solution V1=40 ml

Normality of HCl, N1=0.1 N

To calculate the mass of Na2CO3

$N1\times V1=N2\times V2\\ 0.1\times 40=N2\times 20\\ N2=0.2N$

1 l of solution contain no. of g equivalent=0.2

0.1 l of solution contain no. of g equivalent=0.1*0.2=0.02 g equivalent

No. of g equivalent=sum of (g equivalent of Na2CO3+g equivalent of K2CO3)

No. of g equivalent=mass/equivalent mass

equivalent mass=molar mass/valence

Molar mass of Na2Co3=106 g

Molar mass of K2CO3=138 g

Equivalent mass of Na2CO3=106/2=53 g

Equivalent mass of K2CO3=138/2=69g

Consider mass of Na2CO3=x g

Mass of K2CO3=(1.2-x)g

So we can write

0.02=x/53+1.2-x/69

x=0.59 g=0.6g

How much amount of $CaCO_3$ in gram having percentage purity 50 per cent produces 0.56 litre of $CO_2$ at on heating?

$CaCO_{3}\rightarrow CaO+CO_{2}$

$0.56lt$ of

$CO_{2}=\dfrac{0.56}{22.4}=\dfrac{1}{40}$ mol.

$\therefore \dfrac{1}{40}$ mol of

$CaCO_{3}$ required =

$2.5gm$

$\therefore$ sample is only

$50\%$ pure , thus

$5gm$ of

$CaCO_{3}$ is required.

Lithium exists in nature in the form of two isotopes, $Li-6$ and $Li-7$ with atomic masses $6.0151u$ and $7.0160u$ and the percentage $8.24$ and $91.76$ respectively.
Calculate average atomic mass.

Given:

The atomic mass of

${ Li }^{ 6 }=6.0151\mu$

The atomic mass of

${ Li }^{ 7 }=7.0160\mu$

Percentage abundance of

${ Li }^{ 6 }=8.24$ %

$=0.0824$

Percentage abundance of

${ Li }^{ 7 }=92.70=0.9270$

To calculate the mass of each isotope respectively

average atomic mass= Percentage abundance of isotope

$\times$ atomic mass of isotope 1+Percentage abundance of isotope 2

$\times$ atomic mass of isotope 2

average mass of lithium=

$0.0824\times 6.0151\mu +0.9270\times 7.0160\mu$

$=0.4956\mu +6.5038=6.9994\mu$

The average atomic mass of lithium is

$6.9994\mu$

Calculate the mass percentage composition of the elements in nitric acid (H = 1, N = 14, O = 16).

Nitric acid $HNO_3$ - mol wt. $= 1+14+48$

$=63$ $g$

Mass percentage of hydrogen $= \cfrac {1}{63}\times 100 =1.586%$

Mass percentage of nitrogen $= \cfrac {14}{63}\times 100 =22.22%$

Mass percentage of oxygen $= \cfrac {48}{63}\times 100 =76.19%$

Hence the percentage of $H$ , $N$ and $O$ are $1.58% , 22.22%$ and $76.19%$ respectively.

Write any two differences between mixture and compound. Also write two examples of each.

Compounds:
1. The compound is the chemical combination of elements, bonded together in specific proportion.
2. The compound is a pure substance.
3.Example: water and carbon dioxide.
Mixture:
1.The mixture is the physical combination of substances, bonded together in any proportion.
2.The mixture is an impure substance.
3.Example: sand and water and air (mixture of several gases).

The half life of the nuclide $Rn^{220}$ is $54.5$ sec. Find mass (in kg) of radon is equivalent to $1$ millicurie?

$_{1/2}$ =54.5 sec

Now, t

$_\frac{1}{2}$ =

$\dfrac{0.693}{K}$

Therefore, K=

$\dfrac{0.693}{t_{1/2}}$ =

$\dfrac{0.693}{54.5}$ =

$0.0127 sec$

$^{-1}$

Now , Activity

$=KN$
=

$\cfrac{ K\times weight\quad of\quad element \times N_A}{Atomic\quad weight\quad of\quad element}$

1 millicurie =

$\cfrac{0.0127 \times W\times 6.022 \times 10^{23}} {220}$ dps

$3.7$

$\times 10^{7}$ dps=

$3.476$

$\times 10^{19} \times W$

$W= 1.06 \times 10^{-12}$ g

$W=1.06\times 10^{-15}$ kg

Mass of radon

$=1.06\times 10^{ -15 }$ kg

Calculate the amount of glucose required to prepare $250$ g of $5\%$ solution of glucose by mass.

Percentage of solution=

$5$

Mass of solution=

$250g$

% of solution=

$\cfrac {\text {Mass of solute}}{\text {Mass of solution}}\times 100$

$\therefore$ Mass of solute=

$\cfrac {5 \times 250}{100}= 12.5g$

$4.6 cm^3$ of methyl alcohol is dissolved in $25.2$ g of water. Calculate $\%$ by mass of methyl alcohol.
(Given density of methyl alcohol = $0.7952 gcm^{-3}$ and $C = 12, H = 1, O = 16$ )

Given:

$W_1= 25.2g$

$V_{(CH_3OH)}=4.6cm^3$

$P_{(CH_3OH)}=0.7952 g/cm^3$

Molar mass=

$32g/mol$

$\therefore W_2$ = Density

$\times$ volume

$= 0.7952 \times 4.6$

$= 3.658g$

% by mass=

$\left(\cfrac {3.658}{25.2+3.658}\right)\times 100$

$=\cfrac {3.658}{28.8586}\times 100= 12.67$ %

What is the mass of the precipitate formed when $50$ mL of $16.9\%$ (w/v) solution of $AgNO_3$ is mixed with $50$ mL of $5.8\%$ (w/v) $NaCl$ solution?
[ $Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5$ ]

Moles of

$AgNO_3= \cfrac {50\times 16.9}{100\times 169.8}=0.05$ mole

Moles of

$NaCl= \cfrac {50\times 5.8}{100\times 58.5}=0.05$ mole

$AgNO_3+NaCl \longrightarrow AgCl+NaNO_3$

Mass of

$AgCl$ precipitate= Mole

$\times$ Molar mass

$=0.05 \times 143.5$

$=7.16g$

Hence, 7.16 g AgCl will be precipitated out.

An atom bomb is weighing 1 kg explodes releasing $9 \times 10^{13}$ joule of energy. What percentage of mass is converted into energy?

$E=mc^2$

$\Rightarrow 9\times 10^{13}=m\times \left( 3\times 10^8\right) ^2$

$9\times 10^{13}=m\times 9\times 10^{16}$

$m=\dfrac{9\times 10^{13}}{9\times 10^{16}}=10^{-3}=-.001 \;kg$

$\Rightarrow \%\;mass =\dfrac{0.001\;kg}{1\;kg}\times 100=0.1\%.$

Hence, the answer is

$0.1\%.$

Calculate the mass percent of different elements in calcium phosphate.

$Ca_3\left( PO_4\right)_2$

Total molecular weight

$=310.18\;g/mol$

$\Rightarrow 3Ca=\dfrac{120}{310.18}\times =38.69\%$

$2P=\dfrac{62}{310.18}\times 100=19.99\%$

$80=\dfrac{128}{310.18}\times 100=41.27\%$

Hence, solved.

Calculate the percentage composition in terms of mass of a
solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.

300g of 25% solution contains solute = (300 x 25) / 100 = 75g

& 400g of 40% solution contains solute = (400 x 40) / 100 = 160g

Total solute = 75 + 160 = 235g

As given in the question, total solution =300 + 400 = 700g

Percentage of solute in the final solution = (235 x100) / 700 = 33.5%

Percentage of water in the final solution = 100 - 33.5 = 65.5%

Phosphoric
acid is widely used in carbonated beverages , detergents , toothpastes and
fertilizers . Calculate the mass percentage of H , P AND O in phosphoric acid
if atomis mases are H =1 , P=31 and O = 16

Atomic mass of HYdrogen=

$1 amu$

Atomic mass of Oxygen=

$16 amu$

Atomic mass of phosphorous=

$31 amu$

Total in

$H_3PO_4=3+31+16(4)=98 amu$

% by mass of Hydrogen:

$\cfrac {100\times 1}{98}=1.02$ %

of Oxygen:

$\cfrac {100\times 64}{98}=65.31$ %

of Phosphorous:

$\cfrac {100\times 31}{98}=31.63$ %

Calculate the mass percentage of different constituent elements in urea.

% of element=

$\cfrac {\text {Mass of element}}{\text {Molar mass of element}}\times 100$ %

The molar mass of Urea

$[CO(NH_2)_2]= 60.00 gmol^{-1}$

$\therefore$ % of nitrogen=

$\cfrac {2 \times 14.01}{60.00}\times 100$ %=

$46$ %

% of oxygen=

$\cfrac {1 \times 16}{60.00}=26.6$ %

% of hydrogen=

$\cfrac {4 \times 1}{60.00}= 6.7$ %

% of carbon=

$\cfrac {1 \times 12}{60.00}=20$ %

Rahul Sharma, an intelligent student of class 12 follows the lesson taught in class easily, has friendship with Amit and Krishan who do not follow the lessons easily. Rahul helps Amit and Krishan to understand the concepts and in turn, they are thankful to him. Their friendship is long lasting because both are happy. Rahul feels sense of achievement.
a) Should intelligent students make friends with weak students?Give reason.
b) If you compare this kind of friendship to atoms what kind of bond will be formed.
c) Comment on the strength of such bonding. Name the force that keeps them bonded.
d) Discuss any one property in which these atoms are different from each other.

a) Yes, intelligent students should make friends with weak students because it helps weak people to learn much and intelligent student will get a chance to explain the topic so that he can learn even in better way.

b) On comparing it will look like a co-ordinate covalent bond or dative bond.

c) The strength of dative bond will be very high as more than the covalent bond.

d) Here in this bonded atoms one will have more electrons i.e. contains lone pair of electrons and other atom will have vacant orbital.

A solution is prepared by dissolving 5 g urea in 95 g water. What is the mass percent urea in the solution?

Weight of solution=

$5+95=100g$

$\therefore$ Mass of percentage of urea=

$\cfrac {5}{100}\times 100=5$ %

In a gaseous mixture 2mol of ${ CO }_{ 2 }$ , 1 mol of ${ H }_{ 2 }$ and 2 mol of He are present than determine mole percentage of ${ CO }_{ 2 }$ .

Moles of

${ CO }_{ 2 },{ H }_{ 2 }$

$He$ respectively

$2,1,2$

mole percentage of

${ CO }_{ 2 }=\dfrac { 2 }{ 2+1+2 } =\dfrac { 2 }{ 5 } =0.4\times 100=40$ %

$\therefore$ Mole percentage of

${ CO }_{ 2 }$ is

$40$ %

At $-50^0C$ , liquid $NH_3$ has ionic product is $10^{-30}$ . How many amide $(NH_2^-)$ ions are present per $mm^3$ in pure liquid $NH_3$ ? (take $N_A = 6 \times 10^{23}$ )

$2{ NH }_{ 3 }\longrightarrow { NH }_{ 4 }^{ + }+{ NH }_{ 2 }^{ - }$

Let initial moles of

${ NH }_{ 3 }$ be

$1m$ mole in

$1ml$ .

After dissociation

$\left[ { NH }_{ 4 }^{ + } \right] =x$

$\left[ { NH }_{ 2 }^{ - } \right] =x$

${ x }^{ 2 }={ 10 }^{ -36 }$

$x={ 10 }^{ -18 }$

$\therefore$ Number of ions present per ml of

${ NH }_{ 3 }$

${ N }_{ { NH }_{ 2 }^{ - } }=6\times { 10 }^{ 23 }\times { 10 }^{ -18 }\times { 10 }^{ -3 }$

$=600$

What weight of zinc would be required to produce enough hydrogen to reduce 8.5 g of copper oxide completely into copper?

Solution:-

Reaction of zinc to produce hydrogen-

$Zn + 2 HCl \longrightarrow Zn{Cl}_{2} + {H}_{2}$

Reaction of hydrogen to reduce copper oxide into copper-

$CuO + {H}_{2} \longrightarrow {H}_{2}O + Cu$

From the above reactions-

$1$ moles of

$Zn$ produces

$1$ mole of hydrogen gas and

$1$ mole of hydrogen required to reduce

$1$ mole of

$CuO$ .

Mole of

$CuO$ reduced

$=$ mole of

$Zn$ required

Therefore,

No. of moles of

$CuO$ reduced

$= \cfrac{\text{moles of } CuO}{\text{molar mass of } CuO}$

Given weight of

$CuO = 8.5 \; g$

Molecular weight of

$CuO = 79.5 \; g$

Therefore,

No. of moles of

$CuO$ reduced

$= \cfrac{8.5}{79.5} = 0.107 \text{ mol}$

Therefore,

No. of moles of

$Zn$ required

$= 0.107 \text{ mol}$

Now,

Molecular weight of

$Zn = 65.4 \; g$

$\therefore$ Weight of

$Zn$ required

$= 0.107 \times 65.4 = 6.99 \; g$

Hence

$6.99 \; g$ of

$Zn$ will be required.

Calculate the mass of potassium chlorate required to liberate 6.72 of oxygen at STP.
[Assume: molar mass of potassium chlorate as 122.5 g / mol].

$2 KCl{O}_{3} \longrightarrow 2 KCl + 3 {O}_{2}$

From the above equation,

Amount of

$KCl{O}_{3}$ required to liberate

$\left( 22.4 \times 3 \right) \; L$ of oxygen

$= 2 \times 122.5 \; g$

Therefore,

Amount of

$KCl{O}_{3}$ required to liberate

$6.72 \; L$ of oxygen

$= \cfrac{2 \times 122.5}{22.4 \times 3} \times 6.72 = 24.5 \; g$

Hence,

$24.5 \; g$ of

$KCl{O}_{3}$ is required.

Name the purest form of commercial ion.

Wrought Iron is the purest form of iron. It contains

$0.12$ to

$0.25$ % carbon so, it is the purest form of iron.

44.8 L of $CO_{2}$ at NTP is obtained by heating $x$ g of pure $CaCO_{3}.x$ is:

$CaCO_3\longrightarrow CaO+CO_2$

no. of moles of

$CO_2$ in

$44.8L$ at NTP.

No. of moles

$=\cfrac{\text{Volume of gas}}{\text{Volume at NTP}}$

$n=\cfrac{44.8L}{22.4L}=2$ moles of

$CO_2$ .

$1$ mole of

$CO_2$ is produced by

$1$ mole of

$CaCO_3$ thus,

$2$ moles of

$CO_2$ is produced by

$2$ moles of

$CaCO_3$ .

Amount of

$CaCO_3$ present in

$2$ mole of

$CaCO_3$ ,

$m=2\text{ }mole\times 100 gmol^{-1}$

$n=200g$

By the reaction of carbon and oxygen, a mixture of $CO$ and $CO_{2}$ is obtained. What is the composition(% by mass) of the mixture obtained when 20 grams of $O_{2}$ reacts with 12 grams of carbon ?

1344242_1129362_ans_b00539d4ad524266886acc09940ee129.jpg

Show that ''Law of conservation of mass is fully justified in a balanced chemical equation''.

Law of conservation of mass: The mass of reactants before the reaction is equal to the mass of products after the reaction.

Take the reaction;

$N_2+3H_2\longrightarrow 2NH_3$

Mass of reactants=

$28+3(2)=28+6=34$

Mass of products=

$2(17)=34$

$\therefore$ Mass of reactants before reaction = Mass of products after reaction

$\therefore$ Law of mass conservation is fully justified in a balanced chemical equation.

A solution of ethanol in water is $10%$ by volume. If the solution and pure ethanol have densities of $0.9866\ g/cc$ and $0.785\ g/cc$ respectively, find the percent by weight.

1350699_1139467_ans_5fdc5b5ad8bc41688228440bdbcfb09d.jpg

If the atomic mass of sodium is $23u$ , what will be the atomic mass of sodium ion?

Atomic mass

$=$ Number of neutrons

$+$ Number of protons.

For

$Na,\quad N_n=12,N_p=11$

$23u=12+11$

For,

$Na^+,$ an electron is lost, but the atomic mass remains the same

$(23u)$ as that of

$Na.$

$15.9 \,g$ of copper sulphate and $10.6 \,g$ of sodium carbonate react together to give $14.2 \,g$ of sodium sulphate and $12.3\,g$ of copper carbonate. Which law of chemical combination is obeyed? How?

$CuSO_4+Na_2CO_3\longrightarrow Na_2SO_4+CuCO_3\\15.9\quad+\quad 10.6\quad\quad\quad\quad 14.2\quad +\quad 12.3\\ \quad \quad 26.5g\quad\quad\quad \quad \quad \quad \quad \quad \quad 26.5g$

The total mass of reactants

$=$ total mass of products,

Hence, the law of conservation of mass is obeyed here.

Calculate the percentage loss of mass of hydrated copper [II] sulphate $[ Cu{ SO }_{ 4 }]$ when it is completely dehydrated.
[ $Cu{ SO }_{ 4 }\cdot 5{ H }_{ 2 }O\rightarrow Cu{ SO }_{ 4 }+5{ H }_{ 2 }O$ ]
[At. wts. are Cu = 64, S = 32, O = 16, H = 1].

1352085_1135002_ans_7408b9cf0bdb4a878f64158a2d98b5b7.jpg

A solution of ethanol in water is $10%$ by volume. If the solution and pure ethanol have densities of $0.9866 glcc$ and $0.785 g/cc$ respectively, find the percent by weight.

1350698_1139510_ans_b3258f9d115d48649a411a2d930360d7.jpg

The mole fraction of benzene in a solution in toluene is $0.40$ . Calculate the weight per cent of benzene in the solution.

1342464_1135483_ans_5e9293288d0e48c3a5bd6ba4c0518414.jpg

Define the law of conservation of mass.

The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction.

$27.6$ g ${K_2}C{O_3}$ was treated by a series of reagents so as to convert all of its carbon to ${K_2}Z{n_3}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_2}$ The weight of product is?

1348607_1145567_ans_a427ab4a64984ba6b33590439f07f2ba.jpg

The percent loss in weight after heating a pure sample of potassium chlorate (mol. wt. = 122.5) will be ( ${KCIO }_{3 }\rightarrow KCI +{ O}_{2 }$ )

1348622_1146399_ans_04ec0b46db364ff595871557aaf23d4f.jpg

Phosphoric acid is widely used in carbonated beverages, detergents, toothpastes and fertilizers. Calculate the mass percentages of $H. P$ and $0$ in phosphoric acid if atomis mases are $H = 1, P = 31$ and $0 = 16$ . (Ans. $H - 3.06\%, P - 31.63\%, 0-65.31\%)$

1351496_1142915_ans_a51f5d3f88d943d1b124abd99290eda5.jpg

At STP, 2.8 liters of hydrogen sulphide were mixed with 1.6 liters of sulphur dioxide and reaction occurred according to the equation :
${ 2H }_{ 2 }S(g)+{ SO }_{ 2 }(g)\longrightarrow { 2H }_{ 2 }O(l)+3S(s)$

1353193_1147878_ans_b511adc5873c41abace3eeeaaced053d.jpg

Calculate the molality in the aqueous solution containing $3.0gm$ of urea in $250gm$ of water
( $H=1,N=14,O=16$ )

1300754_1149348_ans_c758714e990b4e2681060335fa3d57bb.jpg

Two samples 'm' and 'n' of slaked lime, $Ca(OH)_2$ , were obtained from two different reactions. The details about their composition are as follows:

$\underline {Sample\ 'm'}$
Total mass: 7 g
Mass of constituent oxygen: 2 g
Mass of constituent calcium: 5 g

$\underline {Sample\ 'n'}$
Total mass: 1.4 g
Mass of constituent oxygen: 0.4 g
Mass of constituent calcium: 1 g

Which law of chemical combination does this prove? Explain.

Slaked lime is

$Ca(OH)_2$

$Sample\ 'm'$	$Sample\ 'n'$
Mass = 7 g	Mass = 1.4 g
Oxygen = 2 g	Oxygen = 0.4 g
Calcium = 5 g	Calcium = 1 g

Proportion of oxygen in sample 'm'

$= \dfrac {2}{7}$

Proportion of oxygen in sample 'n'

$= \dfrac {0.4}{1.4}$

$= \dfrac {2}{7}$

Proportion of calcium in sample 'm'

$= \dfrac {5}{7}$

Proportion of calcium in sample 'n'

$= \dfrac {1}{1.4}$

$= \dfrac {5}{7}$

Proportion of oxygen in both samples is same and that of calcium is also same. This proves the law of constant proportion, i.e. “In a chemical substance the elements are always present in definite proportions by mass”.

Evaporation of water occurs at all temperature
How is eveporation dependent on temperature ?

Evaporation is particularly a surface phenomenon unlike boiling.

During evaporation, the molecule at the surface of liquid take up the heat from the surrounding and gets evaporated to leave the surface of liquid.

* This can happen at any temperature.

* Temperature does affect evaporation as it takes place quickly at high temperatures.

Why does water vapour condense when it reaches high up in the atmosphere ?

Much of the atmosphere is quite cold because sunlight passes through it without getting absorbed much. So, the ground gets hots and water, evaporates when the vapour goes up it reaches coldest parts in the atmosphere and condenses.

At what temperature does water boil ?

It depends on the ambient air pressure.In areas of low pressure-- for instance at high altitudes outdoor(the top of a mountain for instance)--water boils at low tempreture.In a high pressure environment for instance-- pressure cooker, water boils at a higher temperature.

At 1 atmosphere, the standard air pressure at sea level, the boiling temperature of water is 100 degree celcius.

Determine the percentage composition of each atom in the molecule of ${ Na } _ { 2 } { CO } _ { 3 }$ .

The percentage composition of each atom in molecule

$=(no.\ of\ atoms)\times \cfrac {(Atomic\ mass\ of\ element)}{(Molar\ mass\ of element)}\times 100$

$\% Na=\cfrac {2\times 23}{106}\times 100=43.39\%\approx 43.4\%$

$\% C=\cfrac {12}{106}\times 100=11.32\%$

$\% O=\cfrac {3\times 16}{106}\times 100=45.28\%$

Name any two laws of chemical combination.

(i) Law of conservation of mass

(ii) Law of constant proportion

How to calculate the atomic mass?

Atomic mass = no. of proton + no. of neutron

Chlorine has two isotopes of atomic mass units 34.97 and 36.The relative abundance of an isotope is 0.755 and 0.245 respectively. Find the average atomic mass of chlorine.

Average atomic mass

$= (34.97 \times 0.735 + 36.97 \times 0.245)/ (0.735+0.245)$

$= (25.70295 + 9.05765)/0.98$

$= 35.47u$

The atomic number of an element is more important to the chemist than its relative atomic mass. Why?

The atomic number is the number of protons which is equal to the number of electrons in an atom. Now, most of the chemical properties of an element depend upon its valence electrons and electronic configuration. So, atomic number is more important to a chemist, than its relative atomic mass.
Relative atomic mass is the total number of protons and neutrons in an atom, these particles do not take part in most of the reactions, hence are of less importance to a chemist.

The weight of 350 mL of a diatomic gas at $0^0C$ and 2 atm pressure is 1 g. The wt. of one atom is

1352415_1280031_ans_d0e3b8bbed7442bc93af5591c1823b94.jpg

a) Define atomic mass unit.
b) Distinguish between molecular mass and molar mass.
c) Give an example of (i) diatomic, (ii) triatomic molecule of compounds.

(a) Atomic mass unit is the unit of mass used to express atomic and molecular weights, equal to 1/12th of the mass of an atom of carbon-12.

(b) Molecular mass refers to the mass of a molecule, while molar mass refers to the mass of a mole of a substance.

$N_2$ , is a diatomic compound. Ozone,

$O_3$ , is a triatomic compound.

How many grans of concentrated sitric acid $sol^n$ should be used to prepare $250$ mL of $2$ M $HNO_3$ ?
The concentrated acid is $70\%$ $HNO_3$ ?

1331840_1312602_ans_fd14a777b267473599b4102922fe460b.jpg

Calcium chloride, when dissolved in water, dissociates into its ions according to the following equation :
$CaCl_{ 2 (aq)} \rightarrow Ca^{2+}_{aq} +2Cl^- _{aq}$
Calculate the number of ions obtained from $CaCl_2$ when 222 g of its dissolved in water.

$1$ mole→ 111 g

$CaCl_2$

$\therefore$

$222g\,CaCl_2$ is equivalent to

$2$ moles of

$CaCl_2$
Since

$1$ formula unit

$CaCl_2$ gives

$3$ ions, therefore,

$1$ mol of

$CaCl_2$
will give

$3$ moles of ions

$2$ moles of

$CaCl_2$ would give

$3\times2=6$ moles of ions.
No. of ions = No. of moles of ions

$\times$ Avogadro number

$=6 \times 6.022 \times10^{23}$

$=36.132\times10^{23}$

$=3.6132\times10^{24}\,ions$

Distinguish between molecular mass and molar mass.

MOLAR MASS	MOLECULAR MASS
Refer to mass of a mole of a substance	Refer to the mass of molecules
Also known as molecular weight	It determines the mass of single molecule.
SI unit is g/mol to use in higher calculation	Measured in amu
It is defined as the mass of Avogadro number of atom/molecules or compounds	No particular melting point
Measurement given to compounds, atoms or molecules	Determined only in molecules
Less accurate than molecular mass	Accurate to use in higher calculations
Example: Mass of $1$ mole of oxygen is $15.9994$ grams. Therefore the molar mass $=15.9994\,g/mol$	Example : Molecular mass of $Ca (OH)_2=74$ atomic mass units.

atomic mass units

An atom of an element is $10.1$ times heavier than the mass of a carbon atom. What is its mass in a.m.u?

Mass of

$C \ atom = 12 \ amu$

Given; element is

$10.1$ times heavier than the mass of a

$C$ atom.

So, mass of an atom of the element

$= 12 \times 10.1 = 121.2 \ amu$

A gaseous mixture of $H_{2}$ and $N_{2}O$ gas contains 66 mass $\%$ of $N_{2}O$ . What is the average molecular mass of mixture?

1311267_1347010_ans_21f343ef340a4fc7a20d6cbfcdee5af1.jpg

How many amu's are present in ${10^{ - 3}}$ g?

2041520_1378033_ans_e3eb7f960c1b49c58f71b763912b2759.jpg

Express unified atomic mass unit in kg.

1961693_1388684_ans_34c01bb274234e2a800ede0e4cc31f86.jpg

1 L of $CO_2$ is passed over hot coke.When the volume of reaction mixture becomes 1.4 L, the composition of reaction mixture is

1291874_1394639_ans_7632dbb57a924bec924cb87c1e1cd8d2.jpg

Chlorine has two stable isotopes: Cl-35 and Cl-37 with atomic masses 34.96 and 36.95, respectively. If the average mass of chlorine is 35.43, calculate the percentage abundance.

Let x = fraction of Cl-35

Let y = fraction of Cl-37

x + y = 1

So, y = 1 - x

Given, atomic mass of Cl-35 = 34.96; atomic mass of Cl-37 = 36.95; average mass of chlorine = 35.43

34.96 $\times$ x + 36.95 $\times$ y = 35.43

34.96 $\times$ x + 36.95 $\times$ (1 - x) = 35.43 (since, y = 1 - x)

Solving for x gives x = 0.7595

Solving for y gives y = 0.2405

Multiplying with 100 gives percentage abundance:

Percentage abundance of Cl-35 = 0.7595 $\times$ 100 = 75%

Percentage abundance of Cl-37 = 0.2405 $\times$ 100 = 24.05%

Take 30 g of common salt and dissolve it in 70g of water. Find the concentration of a solution in terms of weight percent.

Weight in percent=

$\dfrac{\text{Weight of the solute}}{\text{Weight of the solute+ weight of the solvent}}\times \, 100$

$\quad \quad \quad \quad \quad \quad \quad =\dfrac{30}{30+70}\times \,100 = \dfrac{30}{100} \times \,100=30\%$

$\therefore \text{weight percent}=30\%$

The saturation temperature for 20.7g of $CuSO_4$ soluble in water is_______.

Complete the table given below:
Element Atomic mass Molecular mass Atomicity number
Chlorine 35.5 71 -
Ozone - 48 3
Sulphur 32 - 8
Nitrogen 14 - 2

Element	Atomic mass	Molecular mass	Atomicity number
Chlorine	35.5	71	2
Ozone	16	48	3
Sulphur	32	256	8
Nitrogen	14	28	2

To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.

Mass of solute (sodium chloride) = 36 g

           Mass of solvent (water) = 100 g

           Mass of solution

           = Mass of solute + Mass of solvent

           = 36 g + 100 g = 136 g

Complete the following table:
Element Mass number Atomic Number number of Electrons number of protons number of neutrons
Phosphorous 31 15
Potassium 19 20

Atomic number = Number of protons = number of electrons in a neutral atom
Mass number = Number of protons + Number of neutrons.

Element	Mass number	Atomic Number	number of electrons	number of protons	number of neutrons
Phosphorous	31	15	15	15	31 - 15 = 16
Potassium	19 + 20 = 39	19	19	19	20

What is the active mass of 1L of oxygen gas at N.T.P?

1305260_1607218_ans_9c1fde5eca4a4100b2287a1e1f7d4ecc.jpg

Calculate the formula unit masses of ZnO, $Na_2O$ , $K_2CO_3$ , given atomic masses of Zn = 65u, Na= 23 u, K= 39 u, C= 12 u, and O= 16 u,

1360949_1583724_ans_a398cd5892144bebac3cd3af30e7c6be.jpg

$\displaystyle 3.90\, g$ of a mixture of $Al$ and $Al_2O_3$ , when reacted with a solution of sodium hydroxide, produced $840 \,mL$ of a gas at NTP. Find the composition of the mixture.

$\, W (Al + Al_2O_3) = 3.9\,g$

$Al + 3NaOH \rightarrow Na_3AlO_3 + \dfrac{3}{2} H_2\uparrow$

Volume of

$H_2$ produced

$= 840 \,ml$ (at NTP)

moles of

$H_2$ ------------------

$\dfrac{840}{22400} = 0.0375$

$\displaystyle \because\,\dfrac{3}{2}$ moles of

$H_2$ is produced by

$1$ mole of

$Al$

$\displaystyle \therefore\, 1$ ---------------------

$\dfrac{1}{3/2} = 2/3$ moles of

$Al$

$\displaystyle \therefore \, 0.0375$ -----------------

$\dfrac{2}{3} \times 0.0375 = 0.025$

$\displaystyle \therefore \, Wt.$ of

$Al = 0.025 \times 27 = 0.675\,g$

$\displaystyle \therefore \,$ % of

$Al = \dfrac{0.675}{3.9}\times 100 = 17.3$ %

$\%$ of Aluminium

$100-17.3=82.7\%$

Radium disintegrates at an average rate of $2.24 \times 10^{13} \alpha$ particles per minute. Each $\alpha$ -particle takes up two electrons from the air and becomes a neutral helium atom. After 420 days, helium gas collected was 0.5 mL, measured at $27^0 C$ and 750 mm Hg. Calculate the Avogadro constant.

1610034_1733412_ans_afa3a65722b04e0e942fefac9466fbfa.png

How many years would it take to spend the Avogadro constant of rupees at the rate of $10$ lakh rupees per second?

$\text{Avogadro's number =6.023}\times 10^{23}$

$\text{According to question }$

$\text{if we spend Rs. 10,00,000 in 1 sec,}$

$then,$

$Rs.$

$6.023 \times 10^{23}$

$\text{will be spent in}$

$\dfrac{6.023\times10^{23}}{10^6} sec$

$= 6.023 \times 10^{17}sec.$

$= \dfrac{6.023 \times 10^{17}}{60\times60\times24\times365} years.$

$= 1.91\times10^{10}years$

$3.2\ g$ of a mixture of $KNO_{3}$ and $NaNO_{3}$ was heated to constant weight which was found to be $2.64\ g$ . What is the $\%$ of $KNO_{3}$ in mixture?

$44.22\%$

What is the weight of one atom of H in g (at. wt. of H = $1.008$ )?

$\because$ N atoms of H has wt.

$= 1.008\,g$

$\therefore$ 1 atom of H has wt.

$= \dfrac {1.008}{6.023 \times 10^{23}} = 1.67 \times 10^{-24}\,g$

A mixture of $FeO$ and $Fe_3O_4$ when heated in air to constant weight gains $5$ % in its weight. Find out composition of mixture.

Let weight of

$FeO$ and

$Fe_3O_4$ be a and b g, respectively.

$2FeO + 1/2O_2 \rightarrow Fe_2O_3$

$2Fe_3O_4 + 1/2O_2 \rightarrow 3Fe_2O_3$

$\because$

$144\,g\,FeO$ gives

$160\,g\,Fe_2O_3$

$\therefore$

$a\,g\,FeO$ gives

$160 \times a/144\,g\,Fe_2O_3$

Similarly, weight of

$Fe_2O_3$ formed by b g

$Fe_3O_4 = \dfrac {160 \times 3b}{464}$

Now if

$a + b = 100;$ then

$\dfrac {160 \times a}{144} + \dfrac {160 \times 3b}{464} = 105$

Solving these two equations :

$a = 21.06$ and

$b = 78.94$

Percentage of

$FeO = 21.06\%$ and percentage of

$Fe_3O_4 = 78.94\%$

Does $1$ g of all elements contain nucleons equal to the Avogadro constant? Explain.

1645240_1738245_ans_93234011604e4ba9b6a2eee9ffe01f50.jpg

Find the weight of $H_2SO_4$ in $1200\,mL$ of a solution of $0.4\,N$ strength.

Meq. of

$H_2SO_4 = 0.4 \times 1200 = 480$

$(\because Meq. = N \times V \,in\,mL)$

$\because \dfrac {w}{49} \times 1000 = 480$

$\therefore w = 23.52\,g$

Find the weight of $NaOH$ in its $60$ milli - equivalents.

Meq.

$= \dfrac {w}{E} \times 1000$

$\therefore 60 = \dfrac {w}{40} \times 1000$

$w = 2.4\,g$

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. Also report percentage by weight.

1760804_1742880_ans_437691ac946b436386a65838e1668406.jpg

Name the law of chemical combination :
(a) Which was given by Lavoisier
(b) Which was given by Proust

(a) Law of conservation of mass

(b) Law of constant proportions

A mixture of $NH_3(g)$ and $N_2H_4(g)$ is placed in sealed container at $300 K.$ The total pressure is $0.5 \,atm.$ The container is heated to $1200 K,$ at which time both substances decompose completely according to the equations :
$2NH_3(g) \rightarrow N_2(g) + 3H_2(g)$
$N_2H_4(g) \rightarrow N_2(g) + 2H_2(g)$
After decomposition is complete, the total pressure at $1200\,K$ is found to be $4.5\,atm.$ Find the amount (mole) percent of $N_2H_4(g)$ in the original mixture?

1760633_1742731_ans_8e2e0cf21daa4255bcc81abe1b38df70.jpg

Fill in the blanks :
In water, the proportion of oxygen and hydrogen is _____ by mass.

The atomic mass of

$H$ : 1 u
The atomic mass of

$O$ : 16 u
Proportion of mass of oxygen and hydrogen in

$H_2O$ is given by:

Mass of

$O_2$ : Mass of

$H$ =

$16\times 1:1 \times 2= 8:1$

Name the scientist who gave :
(a) Law of conservation of mass
(b) Law of constant proportions

(a) Law of conservation of mass is proposed by Antoine Lavoisier.

(b) Law of constant proportions is proposed by Joseph Proust.

Explain giving a suitable example:
Law of conservation of mass.

Law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. So, the sum of the masses of the reactants and products remains unchanged.

For example:

$CaCO_3 \rightarrow CaO + CO_2$

100 g 56 g 44 g

Sum of mass of reactants (100 g) = Sum of masses of products [(56 + 44) g]

What is meant by law of conservation of mass? If 12 g of $C$ is burnt in the presence of 32 g $O_2,$ how much $CO_2$ will be formed?

Law of conservation of mass: This law states that "atoms are neither created nor destroyed in a chemical reaction."

This means that the total mass of the products formed in chemical reaction must be equal to the mass of reactants consumed.

$Carbon + Oxygen \rightarrow CO_2$

$Mass \,of \,carbon = 12 \,g$

$Mass \,of \,oxygen = 32 \,g$

$Mass \,of \,CO_2 = ?$

According to the law of conservation of mass,

Sum of masses of reactants = Sum of masses of products

Hence,

$12\,g + 32\,g = 44\,g$

Thus, the amount of

$CO_2$ formed is

$44\,g.$

Who is the father of chemistry? Why?

Robert William Boyle is known as 'Father of Modern Chemistry'.He was an Anglo lrish scientist born in Ireland.He was the first to perform experiments under controlled conditions and publish his researches with elaborate details of procedure, apparatus and observations.Robert Boyle put chemistry on a firm scientific footing transforming it from alchemy into one based on measurements.He defined elements,compounds and mixtures.

Calculate the average atomic mass of hydrogen using the following data:
Isotope $\%$ Natural abundance Molar mass
$^1 H$ $99.985$ $1$
$^2H$ $0.015$ $2$

Average atomic mass is given by the following formula;

$=\dfrac{( Natural\ abundance\ of\ ^1H \times molar\ mass)+ ( Natural\ abundance\ of\ ^2H \times molar\ mass\ of\ ^2H)}{100}$

$= \dfrac{(99.985 \times 1) + (0.015 \times 2)} {100}$

Hence, the average atomic mass of

$H$ comes out to be

$1.00015 \ gm$

Calculate the mass of potassium sulphate required to prepare its $10\%$ solution in $100 \,g$ of water.

Let x g be the mass of potassium sulphate.

We know that:
Mass by mass percentage of a solution

$= \dfrac{Mass \,of \,the \,solute}{Mass \,of \,the \,solution} \times 100$

$\dfrac{10}{100} = \dfrac{x}{100 + x}$

$x = 11.1 \,g$

Define atomic mass.

Atomic mass (or atomic weight) of an element is defined as the relative mass of the atom of the element as compared to the mass of an atom of

$C-12$ isotope.

Calculate the mass of sodium sulphate required to prepare its $20\%$ (mass percent) solution in $100g$ of water.

Let the mass of sodium sulphate required be =

$x\ g$ .

The mass of solution =

$(x+100)g$

$x\ g$ of solute in

$(x + 100) g$ of solution

$20\% =\dfrac{x}{x+100}\times 100$

$20x+2000=100x$

$80x=2000$

$x=\dfrac{2000}{80}$ or

$x=25g$

$25g$ of sodium sulphate will be required to prepare its

$20\%$ solution in

$100g$ of water.

Calculate the mass percent of calcium, phosphorous and oxygen in calcium phosphate $Ca_3(PO_4)_2$ .

Mass percent of calcium

$=\dfrac{3\times ( atomic\ mass\ of\ calcium)}{molecular\ mass\ of\ Ca_2(PO_4)_2}\times 100$

$=\dfrac{120\ u}{310\ u}\times 100=38.71\%$
Mass percent of phosphorous

$=\dfrac{2\times ( atomic\ mass\ of\ phosphorous)}{molecular\ mass\ of\ Ca_2(PO_4)_2}\times 100$

$=\dfrac{2\times 31\ u}{310\ u}\times 100=20\%$
Mass percent of oxygen

$=\dfrac{8\times ( atomic\ mass\ of\ oxygen)}{molecular\ mass\ of\ Ca_2(PO_4)_2}\times 100$

$=\dfrac{8\times 16\ u}{310\ u}\times 100=41.29\%$

Explain fractional atomic mass. What is the fractional mass of chlorine?

Fractional atomic mass is the weighted average of all the naturally occurring isotopes of that element. Fractional mass of chlorine is

$35.5$ .

The total number of protons and neutrons in the nucleus of an atom is known as .

1962526_1823869_ans_1e2d76c9e15e428290816650f6cc76d7.jpg

State the law of conservation of matter or mass.

The law of conservation of mass states that "mass can neither be created nor be destroyed in a chemical reaction".

$\bullet$ The law of conservation of mass was proposed by Lavoisier.

Define law of conservation of mass for a chemical reaction.

Law of conservation of mass states that "mass can neither be created nor destroyed in a chemical reaction".

According to the law of conservation of mass, the mass of the products in a chemical reaction must be equal to the mass of the reactants.

For example,

$N_2 + 3H_2 \rightarrow 2NH_3$

$2\times14 + 3 \times 2 = 2 \times (14+3\times 1)$

$34 = 34$

Therefore, Mass of reactants

$=$ Mass of products

Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.

The mass of an atom is concentrated in its nucleus. It is the sum of proton

$(p)$ and neutron

$(n)$ . The number

$(p+n)$ is called the atomic mass number.

As we know that atom is very small in size, then how do we determine its mass? Therefore, the concept of a reference atom for atomic mass is taken into consideration.

Initially, the hydrogen atom which is the lightest was chosen as the reference atom. The relative mass of the hydrogen atom was accepted as

$1$ . This is how relative atomic masses of various elements were determined in reference to hydrogen.

For example, the mass of one nitrogen atom is fourteen

$(14)$ times that of a hydrogen atom.

Finally, the carbon was selected as a reference atom. The relative mass of the carbon atom was accepted as

$12.00$ . The relative atomic mass of hydrogen compared to the carbon atom becomes

$12 \times 1/12 =1$ .

If bromine atom is available in isotopes $_{ 79 }^{ 35 }{ H } (49.7$ %) and $_{ 81 }^{ 35 }{ H }$ Br $(50.3$ %) Calculate the average atomic mass of bromine atom.

As per data, average atomic mass of bromine is

$(79 \times 49.7$ %

$+ 81 \times 50.3$ %)

$\left( 79\times \frac { 49.7 }{ 100 } +81\times \frac { 50.3 }{ 100 } \right)$

$392.63+407.43$

$800.06$

Average of this means

$\frac{800.06}{2}=400.03.$

The average atomic mass of bromine atom is

$400.03$ .

Give reasons
Actual atomic mass is greater than mass number.

Actual atomic mass is greater than the mass umber because the mass number is a whole number approximation of atomic mass unit.

What is meant by unified mass, u?

Unified mass (u) refers to atomic mass unit (abbreviated as amu). It is defined as 1/12 the mass of an atom of carbon-12. According to latest IUPAC guidelines, we use the symbol u instead of amu to represent masses of atoms.

Name the fundamental law involved in every chemical reaction.

The fundamental law involved in every equation is "the law of conservation of mass". Which states that mass is neither created nor destroyed during a chemical reaction. This means that the mass of the reactants involved in the chemical reaction will always be equal to the mass of the products formed and mass.

What is Avogadro's law? Explain Boyle's law and Charle's law with graph.

(a) Avogadro's law:

$V\propto n$ (at constant

$T$ and

$P$ )

$V=k_{4}n$

$\frac{V}{n}$ = constant

$(k_{4})$

Boyle's law:
Mathematically, Boyle's law is written as,

$V\propto \frac{1}{P}$ (at constant

$T$ and

$n$ )

$V=\frac{k_{1}}{P}$
or,

$PV=k_[1]$
Where,

$P$ = Pressure,

$V$ = Volume,

$k_{1}$ = proportionality constant.
Suppose a gas has volume

$V_{1}$ at pressure

$P_{1}$ . At constant temperature, if pressure extended to

$P_{2}$ and volume becomes

$V_{2}$ , then according to Boyle's law,

$P_{1}V_{1}=P_{2}V_{2}$ = constant.
or,

$\frac{P_{1}}{P_{2}}=\frac{V_{2}}{V_{1}}$
(see the figure 1)
Charle's law:
According to this law, "At constant pressure, the volume of gas increases or decreases by

$\frac{1}{273}$ on increasing or decreasing each degree of temperature".
Mathematically, Charles law can be written as:
Suppose, the volume of gas is

$V_{0}$ at temperature

$0_{0}C$ . If temperature is increased as

$t^{0}C$ then volume becomes

$V_{t}$ , then according to Charles law,

$V_{t}=V_{0}+\frac{V_{0}}{273}*t$
or,

$V_{t}=V_{0}\left ( 1+\frac{t}{273} \right )$
We can calculate the volume of given mass of gas at different temperatures.
(Follow the below figure 2).
For example:
Volume at

$1^{0}C(V_{1})=V_{0}\left ( 1+\frac{1}{273} \right )$
Volume at

$10^{0}C(V_{10})=V_{0}\left ( 1+\frac{10}{273} \right )$

1743481_1829796_ans_acbb9a9f0f834b04bc596271a8cffb05.png

Write a short note on Avogadro's Hypothesis and its applications.

Avogadro's - Hypothesis:

Avogadro's hypothesis states that the volume of a gas is directly proportional to the number of molecules of the gas.

This is represented by the formula,

$\mathrm{V \propto N}$

Applications of Avogadro's hypothesis:
1) In explaining Gay Lussac's law of gaseous volumes.
2) In determining the atomicity of gasses.
3) In determining the molecular formula of a gas
4) In establishing the relationship between relative molecular mass and vapor density.

Mention the signification of mole.

Signification role:

(1) Mole concept gives us a method of calculating the number of atoms present in a given mass of a substance.

(2) It helps in the calculation of the ratio of reactions consumed and products formed quantitatively.

State Avogadro's law.

Avogadro's law states that ''equal volumes of all gases, at the same temperature and pressure, have the same number of molecules ''

What is meant by the relative atomic mass of an element?

The relative atomic mass of an element is the ratio of the mass an atom of the element to one-twelfth of the mass of an atom of carbon.

Relative atomic mass

$=\dfrac{mass\,of\,1atom\,of\,the\,element}{\dfrac{1}{12} \times mass\,of\,1atom\,of\,_6C^{12}\,isotope}$

Calculate the relative molecular mass of carbon dioxide.

Molecular formula of carbon dioxide

$=CO_2$

The atomic mass of Carbon

$= 12$

The atomic mass of Oxygen

$= 16$

$=1 \times$ atomic mass of carbon

$+ 2 \times$ atomic mass of oxygen.

$= (1 \times 12) +(2 \times 16) = 44$

Hence the relative molecular mass of carbon dioxide

$= 44$ .

What is meant by the relative molecular mass of a substance?

The relative molecular mass of a substance is the ratio of the mass of a molecule of the substance to

$\dfrac{1}{12}$ the mass of an atom of Carbon

$_6C^{12}$ isotope.

Why are the atomic masses of most of the elements fractional?

Many elements consist of a mixture of isotopes. Each isotope of an element is a pure substance. The atomic mass of an element is the average relative masses of its various isotopes. While taking this average, the result may appear as a fraction.

Briefly explain laws of chemical combinations.

Laws of chemical combinations are laws that have been established based on various experiments on matter and chemical reactions. Examples of these laws include:

$\text {Law of Conservation of Mass}$ : This law states that matter can neither be created nor be destroyed. In other words, the total mass, that is, the sum of the masses of all reactants and products remains constant.

$\text {Law of Constant Proportions}$ : Joseph Proust, stated that in a chemical substance, the elements are always present in definite proportions by mass.

Calculate the percentage composition of carbon and oxygen in $CO_2$ .

(Given atomic masses: Carbon $= 12$ and Oxygen $= 16$ ).

Mass of carbon dioxide [

$CO_2$ ]

$= 1 (C) + 2(O)$

$= 1(12) + 2(16)$

$= 12 + 32 = 44$

$\%$ mass of carbon

$= \cfrac{12}{44} \times 100 = 21.3\%$

$\%$ mass of oxygen

$= \cfrac{32}{44} \times 100 = 72.7\%$

What do you understand from the statement that the atomic mass of Helium is $4$ ?

Atomic mass of Helium is

$4$ . That is mass of one atom of Helium is

$4$ times of

$1/12^{th}$ mass of carbon atom.

One GAM substance contains Avogadro number of particles in it.
How many particles are there in Avogadro number?

$6.022\times 10^{23}$

One GAM substance contains Avogadro number of particles in it.
Write the number of atoms present in each of the following.
$32g$ Oxygen
(Atomic mass $S=32, O=16, C=12$ )

$2\times 6.022\times 10^{23}$

One GAM substance contains Avogadro number of particles in it.
Write the number of atoms present in each of the following.
$32g$ Sulphur
(Atomic mass $S=32, O=16, C=12$ )

$6.022\times 10^{23}$

Is their any relation between the mass of the number, if the particles are of the same mass.

$Yes,$ their is any relation between the mass of the number, if the particles are of the same mass.

One GAM substance contains Avogadro number of particles in it.
Write the number of atoms present in each of the following.
$32g$ Carbon
(Atomic mass $S=32, O=16, C=12$ )

$\dfrac{32}{12}\times 6.022\times 10^{23}$

Find out the mass of one mole of $CaCo3$ . How many moles of calcium present in $1000\ g\ \ CaCo3$ ?

Mass of

What may be the method of stating the mass of atoms?

The mass of the atom is compared to the mass, of another atom and expressed as a number which shows many times it is heavier than the other atom. The atomic mass of elements are expressed by considering

$1/12$ mass of an atom of carbon

$12$ as one unit.

According to Avagadro's law when the temperature and pressure remain constant on which factor does the volume of gas depend?

Depends on the number of molecules.

Why are atomic mass of some elements are in fractions ?

Atomic masses of most of the elements in atomic mass units involve fractions because the atomic mass of an element is the average of relative masses of its various isotopes.

$Average\ atomic\ mass = f_1M_1 + f_2M_2 +… + f_nM_n$ where f is the fraction representing the natural abundance of the isotope and M is the mass number (weight) of the isotope. While taking the average, the result may appear as a fraction.

Calculate percentage elements present in ammonium chloride $(NH_4Cl)$
$(N=14, H=1, Cl=35.5)$

Molecular mass of

$NH_4Cl =$ (Atomic mass of

$N$ )

$+4\times$ (Atomic mass of

$H$ )

$+$ (Atomic mass of

$Cl$ )

$=(14)+4\times (1)+(35.5)$

$=53.5\ amu$

$\%$ of Nitrogen in

$NH_4Cl =\dfrac {Atomic\ mass\ of\ nitrogen}{Molecular\ mass\ of\ NH_4Cl}\times 100$

$=\dfrac {14}{53.5}\times 100=26.17\%$

$\%$ of Hydrogen in

$NH_4Cl =\dfrac {Atomic\ mas\ of\ Hydrogen\times 4}{Molecular\ mass\ of NH_4Cl}\times 100$

$=\dfrac {1\times 4}{53.5}\times 100=7.48\%$

$\%$ of Chlorine in

$NH_4Cl =\dfrac {Atomic\ mas\ of\ Chlorine}{Molecular\ mass\ of NH_4Cl}\times 100$

$=\dfrac {35.5}{53.5}\times 100=66.35\%$

What is meant by atomic mass and molecular mass?

The mass of an individual atom of an element is known as its atomic mass.

The molecular mass of a substance is the sum total of the atomic masses of the atoms present in that molecule or compound.

Explain the combination capacity of an atom.

The combining capacity of an atom is the number of electron an atom must lose or gain or share to attain a stable configuration. It is also called the valency of an atom.

Complete the flow chart given below, related to one mole of substance.

1762681_1834449_ans_a2bf5a3350c645bebbb16900fac4340c.png

A gas mixture of $3 \cdot 0$ litres of propane and butane on complete combustion at $25^{0}C$ produced 10 litres of $CO_{2}.$ Find out the composition of the gas mixture. $\textbf (M.L.N.R. Allahabad\;1992)$

$C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$

$C_{4}H_{10} + 6 \dfrac{1}{2}O_{2} \rightarrow 4CO_{2} + 5H_{2}O$
Suppose propane

$= x L$
Then buthane

$= (3 - x)L$

$1 L C_{3}H_{8}$ gives

$3L$ of

$CO_{2}$ and

$1 L C_{4}H_{10}$ gives

$4 L$ of

$CO_{2}.$
Hence

$CO_{2}$ produced

$= 3x + 4(3-x)$

$= 12 - x$

$\therefore 12 - x = 10$ (Given)
Hence

$x = 2 L$

$\mathbf{i.e. propane = 2L \;and \;buthane = 1L}$

What is kg-mole? Find out the total number of electrons in a kg-mole of $O_{2}.$

One kg-mol (or one kilo mole) is the molecular mass of the substances expressed in kilogram.

We know that in CGS system one g-mole of substances contain Avogadro no of particles (

$6.023 \times 10^{23}$ ) and in SI system one kg-mol substance contains Avogadro no of particles (

$6.023 \times 10^{26}$ )

Thus one kg-mol of

$O_{2}$ contains

$6.023\times 10^{26}$ molecules. Each oxygen atom contains 8 electrons.

So, total no of electrons =

$6.023 \times 10^{26}\times 2 \times 8=9.6 \times 10^{27}$ no of electrons.

$20 \cdot 0 mL$ of a mixture of oxygen $(O_{2})$ and ozone $(O_{3})$ was heated till ozone was completely decomposed. The mixture on cooling was fond to expand to $21 mL$ . Calculate the percentage of ozone by volume in the mixture.

Decomposition of ozone takes place as follows :

$2O_{3} \rightarrow 3 O_{2}$

2 vol 3 vol

Suppose ozone in the mixture

$= x \;mL.$ Then

$O_{2} = (20 - x) mL$

$2 mL$ of ozone on decomposition give

$O_{2} = 3 mL$

$\therefore x mL$ of ozone on decomposition will give

$O_{2} = \dfrac{3}{2}x = 1 \cdot 5x mL$

$\therefore$ Total volume of mixture after decomposition

$= 1 \cdot 5x + (20 - x)m L$

$= 20 + 0 \cdot 5x$

$\therefore 20 + 0 \cdot 5x = 21$ (Given)

$0 \cdot 5x = 1$ or

$x = 2 \;mL.$

$\therefore$ Percentage of ozone in the mixture

$= \dfrac{2}{20} \times 100 = \mathbf{10}$ %

Why atomic mass is an average value? Explain with a suitable example.

Many elements exist in different forms known as isotopes. They have different atomic masses. So for any specific element, the average atomic mass of all isotopes is considered for different quantitative purposes. It is known as average atomic mass.

For example: chlorine has two isotopes. One isotope with mass of

$34.969 \ amu$ and percentage abundance

$75.77 \%$ whereas other isotope has mass of

$36.965 \ amu$ amd percentage abundance

$24.23 \%$ . Thus average atomic mass will be equal to

$\sum[34.969 \times 0.7577] + [36.965 \times 0.2423] = 35.4536 \ amu$

Why Law of conservation of mass should better be called as Law of conservation of mass and energy?

In nuclear reactions, it is observed that the mass of the products is less than the mass of the reactants. The difference of mass, called the mass defect, is converted into energy according to Einstein equation,

$E = \Delta m c^2$ . Hence we better call it as a law of conservation of mass and energy.

Define atomic mass, and give suitable examples.

Atomic mass is the average mass of atoms of an element, calculated using the relative abundance of isotopes in a naturally occurring element.

It is calculated in the atomic mass unit (amu or u) which is the unit mass for expressing masses of atoms, molecules equal to

$\frac{1}{12^{th}}$ the mass of a single atom of the carbon - 12.

For example the atomic mass of

Magnesium (

$Mg$ ) =

$\text{24.305 u}$ or

$\text{24.305 g/mol}$

Silicon (

$Si$ ) =

$\text{28.0855 u}$ or

$\text{28.0855 g/mol}$

Briefly explain the importance of studying chemistry.

1937059_1866042_ans_86110f5569b84a6ab466c46098015fb8.jpg

Naturally occurring Boron consists of two isotopes whose atomic weights are $10 \cdot 01$ and $11 \cdot$ The atomic weight of natural Boron is $10 \cdot$ Calculate the percentage of each isotope in natural Boron. $\textbf{(M.L.N.R.Allahabad 1994)}$

Suppose the percentage of isotope with atomic weight

$10 \cdot 01 = x.$
Then percentage of isotope with atomic weight

$11 \cdot 01 = 100 -x$

$\therefore$ Average atomic weight

$= \dfrac{10 \cdot 01x + (100 - x) * 11 \cdot 01}{100}$

$= \dfrac{10 \cdot 01x + 1101 - 11 \cdot 01x}{100} = \dfrac{1101 -x}{100}$

$\therefore \dfrac{1101 - x}{100} = 10 \cdot 81$ or

$1101 - x = 1081$
or

$x= 1101 - 1081 = 20$

$\therefore$ % age of isotope with atomic weight

$10 \cdot 01 = \mathbf{20}$ %
% age of isotope with atomic weight

$11 \cdot 01 = \mathbf{80}$ %

The law which states that a chemical compound always contains the same elements combined in a fixed ratio by mass is called _________.

Law of constant composition or law of definite proportions.

Calcium nitrate decomposed on heating according to the equation:
$2Ca(NO_{3})_{2} \rightarrow 2CaO + 4NO_{2} + O_{2}$
The molecular mass of calcium nitrate is 164 u.
Calculate:
a. the volume of nitrogen dioxide $(NO_{2})$ obtained at S.T.P.
b. the weight of calcium oxide obtained when 16.4 g of calcium nitrate is heated to constant weight. [Ca = 40, 0 = 16, N = 14]

$\textbf{(a)}$

$\textbf{Given data:}$

Given mass of calcium nitrate = 16.4 g

Molecular mass of calcium nitrate = 164 u.

$\textbf{Step 1: Calculate the number of moles of calcium nitrate.}$

$Number\ of\ moles\ of\ calcium\ nitrate = \dfrac{Given\ mass\ of\ calcium\ nitrate}{Molecular\ mass\ of\ calcium\ nitrate}$

$Number\ of\ moles\ of\ calcium\ nitrate = \dfrac{16.4}{164}$

$Number\ of\ moles\ of\ calcium\ nitrate = 0.1 mol$

$\textbf{Step 2: Calculate the number of moles of nitrogen dioxide.}$

$\dfrac{Number\ of\ moles\ of\ nitrogen\ dioxide}{Number\ of\ moles\ of \ calcium\ nitrate} = \dfrac{4}{2}$

$\dfrac{Number\ of\ moles\ of\ nitrogen\ dioxide}{0.1} = \dfrac{4}{2}$

$Number\ of\ moles\ of\ nitrogen\ dioxide = \dfrac{4}{2} \times 0.1$

$Number\ of\ moles\ of\ nitrogen\ dioxide = 0.2$

$\textbf{Step 3: Calculate the volume of nitrogen dioxide at S.T.P.}$

$PV=nRT$

$1 \times V = 0.2 \times 0.0821 \times 273$

$$V = 4.48 litre.

$\textbf{Conclusion:}$

Therefore the volume of nitrogen dioxide ( $NO_2$ ) obtained at S.T.P is 4.48 litre.

$\textbf{(b)}$

$\textbf{Step 1: Calculate the number of moles of calcium oxide.}$

$\dfrac{Number\ of\ moles\ of\ calcium\ oxide}{Number\ of\ moles\ of \ calcium\ nitrate} = \dfrac{1}{1}$

$Number\ of\ moles\ of\ calcium\ oxide = number\ of\ moles\ of\ calcium\ nitrate$

$Number\ of\ moles\ of\ calcium\ oxide = 0.1 mol$

$\textbf{Step 2: Calculate the molar mass of calcium oxide.}$

Molar mass of

$CaO$ = Mass of Ca atom + Mass of O atom

Molar mass of

$CaO$ = 40 + 16 = 56 u

$\textbf{Step 3: Calculate the mass of calcium oxide.}$

$Number\ of\ moles\ of\ calcium\ oxide = \dfrac{Mass\ of\ calcium\ oxide}{Molecular\ mass\ of\ calcium\ oxide}$

$Mass\ of\ Calcium\ oxide = Number\ of\ moles\ of\ calcium\ oxide \times Molar\ mass\ of\ Calcium\ oxide$

$Mass\ of\ Calcium\ oxide = 0.1 \times 56$

$Mass\ of\ Calcium\ oxide = 5.6 g$

$\textbf{Conclusion:}$

Therefore, the weight of calcium oxide obtained when 16.4 g of calcium nitrate heated to constant weight is 5.6 g.

When $400$ g of a $20$ % solution by weight was cooled, $60$ g of solute is precipitated. The percentage concentration of the remaining solution is $x \%$ .
The value of $x$ is _______.
(Round figure the answer up to nearest integer value)

The original concentration was

$20$ %, therefore

$80$ g of solute was present originally.

When

$60$ g of solute was precipitated, only

$20$ g of solute is left. The remaining mass of solution

$= 340g$ .

Therefore, the concentration becomes:

$\dfrac{20\times100}{340} = 5.88$ %.

Thus

$x \%=5.88\%$

$\therefore x= 5.88\approx 6$

The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 gm of a copper(II) salt was dissolved in water and added to an excess of potassium iodide solution liberating iodine according to the following equation.
$2Cu^{2+} (aq) + 4I^- (aq)\rightleftharpoons 2CuI(s) + I_2(aq)$
The iodine liberated required $24.5 \:cm^3$ of a $0.100 \:mole \:dm^{-3}$ solution of sodium thiosulphate
$2S_2{O_{3}}^{{2-}}(aq) + I_2 (aq)\rightarrow 2I^-(aq) + S_4{O_{6}}^{{2-}}(aq)$
the percentage of copper, by mass in the copper(II) salt is: $[Atomic \: mass \: of \: copper = 63.5]$

$2Cu^{2+}+4I^{-} \rightleftharpoons2CuI(s)+I_{2}$

$2S_{2}O_{3}^{2-}+I_{2} \rightleftharpoons 2I^{-}+S_{4}O_{6}^{2-}$

m moles of hypo

$=$ m moles of iodine

$\times2$

$=$ m moles of

$Cu^{2+}$ ions.

$=24.5 \times0.1$ m moles.

So, mass of Copper

$=(24.5)(0.1)(10^{-3})(63.5)$ gms.

So,

$\%$ of Copper

$=\cfrac{(24.5)(10^{-3})(0.1)(63.5)}{0.305}\times 100=51\%$

One type of artifical diamond (commonly called $YAG$ for yttrium aluminium garnet) can be represented by the formula $Y_{3}Al_{5}O_{12}$ [ $Y=89, Al=27$ ]. Match element present in List-I with their weight percentages given in List-II.

Weight

$\%$ of

$Y$

$= 267\times \dfrac{100}{594} = 44.95$

Weight

$\%$ of

$Al$

$=135\times \dfrac{100}{594} = 22.72$

Weight

$\%$ of

$O$

$=192\times \dfrac{100}{594} = 32.32$

A sample of clay was partially dried. It was then found to contain $50\%$ silica and $7\%$ water. The original clay contained $12\%$ water. Find the percentage of silica in the original sample (as nearest integer).

(a) For partially dried clay

Per cent of silica is 50%, per cent of water is 43%. Hence, percent of silica + water is

$\displaystyle 50+7=57$ %.

The percent of impurity

$\displaystyle =100-57=43$ %

The ratio of weight of silica to weight of impurity

$\displaystyle =\dfrac {50}{43}$ .

(b) For original sample,

percent of water is 12%, the percent of silica + impurity

$\displaystyle =100-12=88$ .

Let x be the percentage of silica.

$\displaystyle \dfrac {x}{88-x}=\dfrac {50}{43}$

$\displaystyle x=47$ %

Hence, the percentage of silica in the original sample is 47%.

A mixture of $CuSO_4$ and $CuSO_4 \, .5H_2O$ has a mass of $1.245$ g. After heating, it drives off all the water, the mass is only $0.832$ g. What is the mass (in mg) of $CuSO_4.5H_2O$ in the mixture?

Let the mass of

$CuSO_4 = x \: g$

The mass of

$CuSO_4 \, .5H_2O = (1.245 - x)$ g

$\underbrace {CuSO_4 \, .5H_2O }_{ 249.5}\longrightarrow \underbrace {CuSO_4 }_{ 159.5 \: g} +5H_2O(g) \uparrow$

$CuSO_4 \: in \: (1.245 -x) \: g$ of

$\displaystyle CuSO_4 \, .5H_2O=\frac{159.5}{249.5}(1.245 -x) \: g$

Thus,

$\displaystyle \frac{159.5}{249.5}(1.245 -x)+ x = 0.832$

$x=0.1\: g$

The mass of

$CuSO_4 \, .5H_2O= 1.145 \: g$ .

Hence, the mass (in mg)

$CuSO_4.5H_2O$ in the mixture in

$1145$ mg.

The percentage (by mass) concentration of a solution obtained by mixing $300$ g of a $25$ % (solution $I$ ) and $400$ g of a $40$ % (solution $II$ ) by mass is (as the nearest integer):

Mass of solution

$I$

$=$

$300$ g

$\therefore$ Mass of solute in solution

$I$

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

Mass of solution

$II$

$=$

$400$ g

Mass of solute in solution

$II$

$=\displaystyle\frac{40\times 400}{100}=160$ g

Now, total mass of solute in solutions after mixing

$= 75+160 = 235$ g

Total mass of solution after mixing

$= 300+400 = 700$ g

% by mass

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

So, the answer is

$34\%$ .

A solution of a specific gravity $1.6\ g\ {mL}^{-1}$ is $67\%$ by mass. The $\%$ by mass of the solution of the same acid, if it is diluted to the specific gravity $1.2\ g\ {mL}^{-1}$ is (as nearest integer) :

Assume

$100$ mL of solution of

$67\%$ by mass of specific gravity

$1.6$

$g\: {mL}^{-1}$ .
Mass of solution

$= 100\times 1.6= 160\:g$ .
Mass of solute

$=\displaystyle\dfrac{67\times 160}{100}=107.2\:g$ .
Now, consider

$X\:g\:{H}_{2}O$ is added to it.
Mass of new solution

$=(160+X)\:g$ ...(i)
Also,

$X\: g\:{H}_{2}O$ means

$X$ mL of

${H}_{2}O$ and, thus,
Volume of new solution

$=$

$(100+X)$
Using specific gravity of the solution, the mass of new solution

$= (100+X)\times 1.2= (120+1.2X)\:g$ ...(ii)
By Eqs. (i) and (ii), we get

$160+X=120+1.2X$

$X$

$=200$ mL or

$200$ g

$\%$ by mass of new solution

$=\displaystyle\dfrac{107.2}{160+200}\times 100=29.78\%$ .
So, the answer is

$30$ .

A regular co-polymer of ethylene and vinyl chloride contains alternative monomers of each type. The mass percent of ethylene in this co-polymer is (as the nearest integer) :

The molar masses of ethylene ad vinyl chloride are 28 g/mol and 62.5 g/mol respectively.
Since the ratio of ethylene and vinyl chloride in the polymer is 1:1, the mass percent of ethylene in the co polymer is

$\displaystyle \frac {\text {Molar mass of ethylene}}{\text {Molar mass of ethylene + molar mass of vinyl chloride}} \times 100 =\frac {28}{28+62.5} \times 100 = 31$ %

Cytochrome c is an iron-containing enzyme found in the cells of all aerobic organisms. If cytochrome c is $0.43\%$ $Fe$ (atomic mass = 56 amu) by weight, then the minimum molecular weight (Kg.mol $^{-1}$ ) is $6.5X$ . Then $X$ is:

Let the molar mass be

$=M$ .

$0.43\%$

$Fe$ has molar mass

$=100\ g$ .

Then,

$56$ g

$Fe$ has molar mass

$=\displaystyle\frac{100\times 56}{0.43}=13023.26 = 13 \:Kg$

$mol^{-1}$ .

Hence,

$X=2$ .

A polystyrene having the formula ${Br}_{3}{C}_{6}{H}_{3}{({C}_{8}{H}_{8})}_{n}$ , was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air.

If it was found to contain $10.46\%$ of bromine by mass, the value of $n$ is:

% of Bromine =

$\dfrac{\text{mass of Bromine} }{\text{mass of compound}}\times 100$

Molar mass of

$Br_3C_6H_2(C_8H_8)n$

$=3 \times 80 +12\times 6+2\times 1+(12\times 8+1\times 8)n$

$=314+104n$

$\therefore 10.46=\dfrac{3\times 80}{314+104n}\times 100$

$\therefore 0.1046(314+104n)=240$

$\therefore 10.87n = 240-32.84$

$\boxed {n=19.05 \approx 19}$

When $400$ g of a $20\%$ solution by mass was cooled, $50$ g of solute precipitated. The percentage (by mass) concentration of the remaining solution is (as the nearest integer) :

Mass of solution

$=$

$400$ g
Mass of solute

$=\dfrac{20\times 400}{100}=80$ g
Mass of solute precipitated

$=$

$50$ g
Mass of solute remained in solution

$= 80-50 = 30$ g
And, mass of solution remained

$=$

$400 -50$

$=$

$350$ g

$\%$ (by mass) concentration

$=\dfrac{30}{350}\times 100=8.57\%$
So, answer is

$9$ .

To prepare $100$ g of a $92$ % by weight solution of $NaOH$ , how many grams of $H_2O$ is needed?

Weight of

$NaOH=\left (\dfrac {92 \:g}{100\: g\: solution}\right )$

$(100\: g\: solution)$

$=92\ g\: NaOH$
Weight of

$H_2O=100-92=8.0\ g\: H_2O$

One mole of an ideal gas at standard temperature and pressure occupies 22.4 L(molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen ? (Take the size of hydrogen molecule to be about 1).
Why is this ratio so large ?

Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10^-10 m
Volume of hydrogen atom = (4/3) π r³
= (4/3) × (22/7) × (0.5 × 10^-10)³
= 0.524 × 10^-30 m³

Now, 1 mole of hydrogen contains 6.023 × 10²³ hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms, V_a = 6.023 × 10²³ × 0.524 × 10^–30
= 3.16 × 10^–7 m³

Molar volume of 1 mole of hydrogen atoms at STP,
V_m = 22.4 L = 22.4 × 10^–3 m³

$\therefore \dfrac{V_m}{V_a}=\dfrac{22.4 \times 10^{-3}}{3.6 \times 10^{-7}}=7.08 \times 10^4$

Hence, the molar volume is 7.08 × 10⁴ times higher than the atomic volume.

The ratio is so large because inter-atomic separation in hydrogen gas is large.

1 litre of $N_2$ and $\cfrac{7}{8}$ litre of $O_2$ at same temperature and pressure were mixed together. The ratio of the masses of two gases in the mixture is _______.

1756853_452982_ans_943a16e429fe4f2b8f38219fc5b68558.png

A $0.24\ g$ sample of compound of oxygen and boron was found by analysis to contain $0.096\ g$ of boron and $0.144\ g$ of oxygen. Calculate the percentage composition of the compound by weight.

Mass of the given sample compound = 0.24g

Mass of boron in the given sample compound = 0.096g

Mass of oxygen in the given sample compound = 0.144g

% composition of compound = % of boron and % of oxygen

Therefore % of boron = (mass of boron)/(mass of the sample compound) x 100

= 40%

Therefore % of oxygen = (mass of oxygen) / (mass of the sample compound) x 100

= 60%

The relative atomic mass of an element A is $16.2$ there are two isotopes $_{8}^{16}A$ and $_8^{18}A$ of the element. Calculate the percentage of these two isotopes present in the element.

Data:-

Relative atomic mass=

$16.2$

Atomic mass of

$^{16}_8A=16$

Atomic mass of

$^{18}_8A=18$

Formula:-

Let percentage of Isotope

$^{16}_8A$ be

$\longrightarrow x$

$\therefore$ Percentage of Isotope

$^{18}_8A\longrightarrow (100-x)$

Relative atomic weight=

$\left[Atomic\quad weight\quad of\quad ^{16}_8A\times abundance + Atomic\quad weight\quad of\quad ^{18}_8A\times abundance\right]/100$

$\therefore 16.2=\cfrac {16\times x+18\times (100-x)}{100}=\cfrac {16x\times 1800-18x}{100}$

$\Rightarrow 16.2\times 100=1800-2x$

$\Rightarrow 1620=1800-2x$

$\Rightarrow 2x=1800-1620$

$\Rightarrow 2x=180$

$\therefore x=90$

$100-x=10$

Percentage of

$^{16}_8A$ is

$90$ %

Percentage of

$^{18}_8A$ is

$10$ % .

In an experiment $1.288\ g$ of copper oxide was obtained from $1.03\ g$ of copper and $0.258\ g$ of oxygen. Calculate the ratio of copper and oxygen in the sample?

$2Cu + O_2 \rightarrow 2CuO$

$1.03g$

$0.258g$

$1.288g$

Therefore, since

$Mass \ of \ reactants= Mass \ of \ products$ . It is following the law of conservation of mass.

$\cfrac {Ratio \ of \ Copper}{Ratio \ of \ Oxygen}= \cfrac {1.03}{0.258}= \cfrac {3.99}1 = 4:1$

In a Gobar gas plant, gobar gas is formed by bacterial fermentation of animal refuse. It mainly contains methane and its heat of combustion is $-809\ kJ$ ${mol}^{-1}$ according to the following equation:
${ CH }_{ 4 }+2{ O }_{ 2 }\longrightarrow { CO }_{ 2 }+2{ H }_{ 2 }O;\Delta H=-809\ kJ\quad$
How much gobar gas would have to be produced per day for a small village of $50$ families, if it is assumed that each family requires $20,000\ kJ$ of energy per day? The methane content in gobar gas is $80$ % by mass.

Energy consumption of

$50$ families per day

$=50\times 20,000kJ=1\times {10}^{6}kJ$

$809kJ$ of energy is obtained by burning methane

$=16g$

$=\cfrac{16}{809}\times {10}^{6}=1.98\times {10}^{4}g$

$=19.8Kg$
Since, methane content in gobar gas is

$80$ % by mass, hence, the mass of gobar gas needed

$=\cfrac { 100 }{ 80 } \times 19.8=24.75kg\quad$

$One\ gram$ of charcoal adsorbs $100\ mL$ of $0.5\ M\, CH_3COOH$ to form a monolayer and thereby the molarity of acetic acid is reduced to $0.49\ M$ . Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal $= 3.01 \times 10^2\, m^2/gm$ .

Moles of

$CH_3COOH$ absorbed

$=$ Initial moles -Final moles.

$=0.5\times 100\times 10^{-3}-0.049\times 100\times 10^{-3}$

$=0.05-0.049=0.001$

No. of molecules of

$CH_3COOH$ absorbed

$=0.001\times 6.02\times 10^{23}$

$=6.02\times 10^{20}$

Surface area

$= 3.01\times 10^2m^2$ (given)

Surface area of charcol absorbed by in each molecule of

$CH_3COOH=\cfrac{3.01\times 10^2}{6.02\times 10^{20}}=5\times 10^{-19}m^2$

A commercial sample (2.013 g) of NaOH containing $Na_2CO_3$ as an impurity was dissolved to give 250 mL of solution. A 10 mL portion of the solution required 20 mL of 0.1 N $H_2SO_4$ solution for complete neutralisation of $NaOH$ . Calculate the percentage by weight of $Na_2CO_3$ in the sample.[Write up to 2 decimal places]

Given, Mass of Sample

$=2.013g$

Volume of

$250ml$ solution of sample used by neutralization

$=10ml$

Voluem of

$H_2SO_4$ required for it

$=20ml$

Normality of

$H_2SO_4=0.1N$

At equivalence,

No. of gm equivalents of

$NaOH =$ No. of gm equivalents of

$H_2SO_4$

$[\because H_2SO_4$ reacts only with

$NaOH$ in sample solution not with salt

$Na_2CO_3]$

$\Rightarrow \underset {NaOH}{N_1V_1}=\underset {H_2SO_4}{N_2V_2}$

$N_1=\cfrac {N_2V_2}{V_1}=\cfrac {0.1\times 20}{10}=0.2N$

$\therefore$ Normality of

$NaOH$ in solution

$=0.2N$

$\Rightarrow$ Molarity of

$NaOH$ in solution=

$\cfrac {0.2}{nfactor}$

$=\cfrac {0.2}{1}=0.2M$

$[\because$ Here

$n factor$ for

$NaOH=1$ as acidity of

$NaOH=1]$

$\Rightarrow$ Molarity of

$NaOH=0.2M$

That means,

$0.2$ moles of

$NaOH$ dissolved in

$1000ml$ solution

$\therefore$ No. of moles of

$NaOH$ in

$250ml$ solution is

$\cfrac {0.2\times 250}{1000} = 0.05$ moles

Now, for mass calculation,

$1$ mole

$NaOH=40g$

$\Rightarrow 0.05$ mole

$NaOH=40\times 0.05$

$=2g$

$\therefore$ Mass of

$NaOH=2g$

$\Rightarrow$ Mass of impurity

$Na_2CO_3=$ Total mass - mass of

$NaOH$

$=2.013g- 2g$

$=0.013g$

Now, % calculation

Mass % of

$Na_2CO_3=\cfrac {Mass\ of\ Na_2CO_3}{Total\ mass}\times 100$

$=\cfrac {0.013}{2.013}\times 100$

$=0.6458$ %

A gas cylinder can hold 1 Kg of hydrogen at mom temperature and pressure:
i) Find the number of moles of hydrogen present
ii) What weight of $CO_{2}$ can the cylinder hold under similar conditions of temperature and pressure? ( $H=1,\ C=12,\ O=16$ )
iii) If the number of molecules of hydrogen in the cylinder is $X$ , calculate the number of $CO_{2}$ molecules in the cylinder under the same conditions of temperature and pressure
iv) State the law that helped you to arrive at the above result.

According to Avogadro's law

at constant temperature, pressure & volume; number of moles of two gases become equal.

$n_1=n_2$ or

$n_{H_2}=n_{CO_2}$

(i) Number of moles in

$1kg$ of

$H_2= \cfrac {1000}{2}g= 500$ moles

Thus,

$n_{CO_2}=n_{H_2}=500$

(ii) Mass of

$CO_2$ in

$500$ moles=

$500 \times 44=22000$ g or

$22kg$

(iii) Number of hydrogen molecule in

$500$ moles be

$X$

$X=500 \times M_a$ molecules

$X= 500 \times 6.022 \times 10^{23}$ molecules

Thus number of

$CO_2$ molecules in

$500$ moles of

$CO_2$ will be

$X= 500 \times M_a$ molecules

$X= 500 \times 6.022 \times 10^{23}$ molecules .

(iv) The law is Avogadro's Law .

$10g$ of a sample of $Ba(OH)_2$ is dissolved in $10mL$ of $0.5N$ $HCl$ solution. The excess of HCl was titrated with $0.2N$ $NaOH$ . The volume of $NaOH$ used was $20cc$ . Calculate the percentage of $Ba(OH)_2$ in the sample.

$Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O$

Mili equivalents of

$HCl$ initially =

$5$

Mili eq of NaOH consumed = Mili eq HCl consumed = 4

Mili eq of

$HCl$ finally = 1= Mili eq of

$Ba(OH)_2$

Now the wt of

$Ba(OH)_2$

$=\dfrac{10^{-3}}{171\times 2}$

$Wt = 3\times 10^{-3}$

$\% in sample\approx .03\%$

Calculate the mass percentage of benzene $\left( {C_6 H_6 } \right)$ and carbon tetrachloride $\left({CCl_4}\right)$ if $22\ g$ of benzene is dissolved in $122\ g$ of carbon tetrachloride.

In general Mass % of

$X=\left[\cfrac {\text{Mass of X}}{\text{Mass of total mixture}}\right]\times 100$

Mass of the total mixture=

$22+122=144g$

Mass % of benzene=

$\left(\cfrac {22}{144}\right)\times 100= 15.28$ %

Mass % of

$CCl_4=\left(\cfrac {122}{144}\right)\times 100= 84.72$ %

How many g of a $5.0\%$ by weight NaCl solution are necessary to yield $3.2$ g NaCl?

$5$ % $kg$ weight $NaCl$ means $= \cfrac {5 \ g \ NaCl}{100 \ gm (water+ NaCl)}$

i.e, $5 \ g \ NaCl$ in $95 \ g$ water

then, $3.2 \ g \ NaCl$ in $x \ g$ water

Weight of solution $= 60.8 +3.2 = 64 \ g$ solution

How many $Cl$ atoms can you ionize in the process $Cl\rightarrow {Cl}^{+}+e$ by the energy liberated for the process $Cl+e\rightarrow {Cl}^{-}$ for one Avogadro number of atms. Given $IP=13.0eV$ and $EA=3.60eV$ . Avogadro number $=6\times {10}^{23}$ .

Ionization potential is the energy required to remove an electron from the outermost shell of an atom.

$\therefore Cl\longrightarrow Cl^{+}+e^{-}$ ;

$E=13.0$

$eV$

Electron affinity is the property of an atom to attract an electron.

$\therefore Cl+e^{-}\longrightarrow Cl$ ;

$E=-3.6$

$eV$

$\therefore$ Moles of Cl atoms that can be ionized=

$\cfrac {3.6\quad eV}{13.0\quad eV}=0.277$ moles

No. of atoms=

$0.277\times 6.022\times 10^{23}$

$1.667\times 10^{23}$ atoms

An element has average atomic mass $20.426u$ it has $2$ isotopes with masses $22.1824$ and $19.9574$ . Calculate the abundance of its isotopes $X-22$ and $X-20$ .

We know that,

$\begin{array}{l}20.426 = \frac{{\left( {22.1824 \times x} \right) + \left( {19.95} \right)\left( {100 - x} \right)}}{{100}}\\ \Rightarrow x\left( {22.1824 - 19.95} \right) = 2042.6 - 1995\\ \Rightarrow x\left( {2.2324} \right) = 47.6\\ \Rightarrow x = \frac{{47.6}}{{2.2324}} = 21.32\\x - 22 \to 21.32\% \\x - 20 \to (100 - 21.32) = 78.68\% \end{array}$

Hence the abundance are

$21.32\%$ and

$78.68\%$ respectively

$0.2g$ of an organic compound on analysis given $0.147g$ of carbondioxide, $0.12g$ of water and $74.6 c.c$ of nitrogen at STP. Calculate the weight percentages of constituents.

$C_XH_YN_A + \dfrac{Y}{2}O_2 \longrightarrow XCO_2 + \dfrac{Y}{2}H_2O+\dfrac{A}{2}N_2$

given

$.147gm$ of

$CO_2$ means

$mol = \dfrac{.147}{44} = 3.3\times 10^{-3}$

means

$C$ has the same

$mol = 3.3\times 10^{-3}mol$ so

$wt = 12\times 3.3\times 10^{-12}gm$

and oxygen has the

$mol = 2\times 3.3\times 10^{-3}$ so

$wt= 6.6\times 10^{-3}\times 16gm$

given

$.12gm$ water means the

$mol =\dfrac{.12}{18} = 6.6\times 10^{-3}$

moles of

$H_2 = 6.6\times 10^{-3}$ so

$wt = 6.6\times 10^{-3} \times 1gm$

Moles of oxygen

$= \dfrac{6.6\times 10^{-3}}{2} = 3.3\times 10^{-3}$ so

$wt = 3.3\times 10^{-3}\times 16gm$

given

$74.6cc$ nitrogen means mole

$= \dfrac{74.6}{22400} = 3.3\times 10^{-3}$ so

$wt = 3.3\times 10^{-3}\times 14gm$

Weighr % of C

$= \dfrac{36\times 10^{-3}}{0.2}\times 100=18 \%$

Weight % of H

$= \dfrac{6.6\times 10^{-3}}{0.2}\times 100=3.3\%$

Weight % of O

$= \dfrac{48\times 10^{-3}}{0.2}\times 100=24\%$

Weight % of N

$= \dfrac{42\times 10^{-3}}{0.2}\times 100=21\%$

A solution contains 40 g of common salt in 320 g of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Mas of common salt (solute)=40 g

Mass of water (soluted) =320 g

mass of solution=

$320+40$

=360 g

concentration of solution=

$\dfrac{M_{mass} of solute}{mass of solution}\times100$

$\dfrac{40}{360}\times 100=11.11%$

Hence, the concentration of the solution is 11.11%

Which would be larger : an atomic mass unit based on the current standard or one based on the mass of a Be- 9 atom set at exactly 9 amu ?

$Be$ would be lighter on this scale than it is on the

$C-12$ scale.

Reason:- The current system takes

$1amu$ as

$\cfrac {1}{12}{th}$ that of

$C^{12}$ atom. The

$C^{12}$ atom consists of

$6$ protons and

$6$ neutrons, so the average mass would be the average of proton and neutron. Now if a

$Be^9$ atom is considered, it has

$4$ protons and

$5$ neutrons, making the average tilted towards the neutron side, thereby giving the value of

$(amu)$ as slightly more than that given by the

$C^{12}$ atom. Hence, the one based on

$Be^9$ atom is larger.

Density of dry air (only $N_{2}$ & $O_{2}$ ) is $1.146\ g\ litre^{-1}$ at $740\ mm$ and $300 K$ . Calculate composition of air by weight assuming ideal nature.

Let us assume the volume to be

$1$ litre

$\therefore 1.146=\cfrac {m}{1}$

$m=1.146g$

Now,

$PV=nRT$

$740mm=0.97$ atm

$\therefore 0.97\times 1=\cfrac {m_{ mix }}{M_{mix}}\times 0.0821\times 300$

$\therefore M_{ mix }=\cfrac {1.146\times 0.0821\times 300}{ 0.97 }$

$\therefore M_{ mix }= 29.1$

But

$M_{ mix }=X_{ O_2 }\times M_{O_2 }+X{ N_2 }\times M_{ N_2 }$

where

$X_{ O_2 }$ &

$X_{ N_2 }$ are the mole fractions of

$O_2$ and

$N_2$ respectively.

$\therefore 29.1=32\times X_{ O_2 }+28\times X_{ N_2 }\longrightarrow (i)$

and

$X_{ O_2 }+X_{ N_2 }=1\longrightarrow (ii)$

Solving (i) & (2), we get,

$X_{ O_2 }=0.275$ and

$X_{ N_2 }=0.725$

Weight % of

$O_2=\cfrac { 0.275\times 32 }{ (0.275\times 32)+(0.725\times 28) }\times 100=30.24$ %

$\therefore$ Weight % of

$N_2= 100-30.24=69.76$ %

$\therefore$ Composition of air is

$O_2=30.24$ %

$N_2=69.76$ % {by weight}

Compute the number of ions present in $5.85$ g of sodium chloride.

1870370_1141531_ans_108b6c63fa4842f1891ae2d576a6e691.jpg

A sample of lead weighing $1.05g$ was dissolved in a small quantity of nitric acid to produce an aqueous solution of ${Pb}^{2+}$ and ${Ag}^{+}$ (which is present as impurity). The volume of the solution was increased to $300ml$ by adding water, a pure silver electrode was immersed in the solution and the potential difference between this electrode and a standard hydrogen electrode was found to be $0.503V$ at ${25}^{o}C$ . What was the % of $Ag$ in the lead metal? (Given : ${ E }_{ ({ Ag }^{ + }/Ag) }^{ o }=0.799V$ . Neglect amount of ${Ag}^{+}$ converted to $Ag$ )

Solution:

The mass fraction of silver is :

$\omega =\dfrac { m(Ag) }{ m(\text{sample}) }$

where m(sample) is 1.05g . Lets find the concentration of silver in solution. For this, we can use Nernst equation:

$E={ E }^{ 0 }+0.0591\times ln[A{ g }^{ + }]$

$0.503=0.799+0.0591\times ln[A{ g }^{ + }]$

$ln[A{ g }^{ + }]=\quad -5.01$

$[A{ g }^{ + }]=6.68\times { 10 }^{ -3 }mol{ L }^{ -1 }$

The number of the mioles of

$A{ g }^{ + }$ in 300ml solution is :

$n(Ag)=[{ Ag }^{ + }]\times V\quad =6.68\times { 10 }^{ -3 }\times 300\times { 10 }^{ -3 }\quad =2.00\times { 10 }^{ -3 }$ mol

The mass of the silver is :

$m(Ag)\quad =n(Ag)\times M(Ag)=2.00\times { 10 }^{ -3 }\times 107.8682\quad =0.216g$

Mass fraction of the silver is :

$\omega =\dfrac { 0.216 }{ 1.05 } =0.21\quad or\quad 21$ %

Hence answer is 21% of Ag

Average atomic weight of an element $M$ is $51.7$ . If two isotopes of $M,\ M^{50}$ and $M^{52}$ are present, then calculate the percentage of occurrence of $M^{50}$ in nature.

Average Atomic weight =

$F_{1}M_{1}+F_{2}M_{2}+...F_{n}M_{n}$

$F_{1},F_{2},F_{3}...F_{n}$ are percentage fraction.

$M_{1},M_{2},M_{3}...M_{n}$ are molecular weight of isotope.

$51.7=x\times (50)+(1-x)52$

$51.7= 50x+52-52x$

$-2x=-0.3$

$x=\frac{0.3}{2}$

$x= 0.15$

Abundance of

$M^{50}$ in nature is 15%

and Abundance of

$M^{52}$ in nature is 85%

The balanced equation for a reaction is given below
$2X + 3Y \to {\text{4I + m}}$
When 8 moles of x react with 15 moles of y, then
i) Which is the limiting reagent?
ii) Calculate the amount of products formed.
iii) Calculate the amount of excess reactant left at the end of the reaction.

$2X + 3Y \to 4I + m.$

$8$ mole of

$x$ requires

$\to \frac{3}{2} \times 8 = 12$ mole of

$Y$

$15$ moles of

$Y$ requires

$\to \frac{2}{3} \times 15 = 10$ moles of

$X$

but only

$8$ moles of

$X$ are given so.

$(i)$ .

$X$ is limiting reagent (which decides product)

$(ii)$ .

$8$ moles of

$X$ gives

$16I$ and

$4m$ .

$(iii)$ .

$3$ moles of

$Y$ remains unreacted.

Samples of $O_2, N_2$ and $CO_2$ under the same conditions of temperature & pressure contain the same number of molecules represented by $X$ . The molecules of oxygen occupy $V$ litres and have a mass of $8\ g$ . Under the same conditions of temp and press., what is the volume occupied by:
i) $X$ molecule of $N_2$
ii) $3X$ molecules of $CO_2$
iii) What is the mass of $CO_2$ in grams.
iv) In answering the above questions, whose law has been used.

2106728_1158810_ans_16ec073b490247e9b6abcaf8617dcec5.jpg

A mixture of CO of $CO_2$ is found to have a density of $1.5gm/ L$ at $30^oC$ & 730 torr What is the composition of the mixture.

1353204_1153591_ans_3983d95e590b45b584580d802b3edbd9.jpg

One of the methods of preparation of per disulphuric acid, $H_2S_2O_8$ involve electrolytic oxidation of $H_2SO_4$ at anode $(2H_2SO_4 \rightarrow H_2S_2O_8 +2H^+ +2e^-)$
with oxygen and hydrogen as by-products. In such an electrolysis, 9.722 L of $H_2$ and 2.35 L of $O_2$ were generated at STP. What is the weight of $H_2S_2O_8$ formed ?

Given that

Volume of

$H_2$ produced

$V_{H_2}=9.722\,lit$

Volume of

$O_2$ produced,

$V_{O_2}=2.35\,lit$

Now cell reaction:-

Anodic reaction:

$2H_2SO_4\rightarrow H_2S_2O_8+2H^++2e^-$

$2H_2O\rightarrow 4H^+ +O_2(g)+4e^-$

Cathodic reaction : Refer image (1)

Now,

$\dfrac{x}{2F}=\dfrac{9.722lit}{22.4lit}\Rightarrow x=\dfrac{9.722}{22.4}\times 2F=0.868F$

Note: Electroon condumed in cathod will be equal to electron generated at

Anode.

Again,

Refer image.2

Now,

$\dfrac{y}{4F}=\dfrac{2.35lit}{22.4lit}\Rightarrow y=\dfrac{2.35lit}{22.4lit}\times 4F=0.4196F$

$\left.\begin{matrix}change\,(electron)produced\,from\,H_2SO_4\\ reaction,q_{H_2SO_4}\end{matrix}\right\}=q_{H_2}-q_{o_2}$

$=(0.868-4196)F$

$=0.4484F$

Again

Refer image (3)

Now,

$\dfrac{Z}{194g}=\dfrac{0.4484F}{2F}\Rightarrow Z=\dfrac{0.4484F}{2F}\times 194g=43.49g$

$\therefore$ weight of

$H_2S_2O_8$ formed

$\approx 43.49g$

1426638_1346374_ans_b2067088a0114400b4c2bd740ead6c67.jpg

In a solution of urea, the mole fration of urea is $0.2$ . Calculate the mass $\%$ of solute

0.2 moles of solute (urea) =0.2 X molar mass of urea = 0.2 X 60=12 grams

0.8 moles of solvent (water) = 0.8 X molar mass of water = 0.8 X 18= 14.4 grams.

Mass of solution = Mass of solute + Mass of solvent

=12 + 14.4

26.4 grams

Volume =mass/density

=26.4/d mL =26.4/1000 dL

Molarity =0.1/26.4/1000 d

1/10 X 1000d/26.4

Molarity =100d/26.4

If density=0.21gm/mL putting this value in the above equation we get.

Molarity=100X0.21/26.4=0.79 M

if d=0.21 gm/mL Its volume=26.4/1000 X0.21

volume=0.12 L =0.12 X1000= 120 ml

Percentage of solution= mass/volume of solution

=26.4/120

=0.22%

A mixture of $FeO$ and $Fe_{3}O_{4}$ when heated in air to a constant weight gains $5\%$ in its weight. Find the composition of initial mixture.

1680277_1730977_ans_1a50ec58fb534dabaf67d81eae4c04de.png

A mixture containing $NaHCO_{3}$ and $Na_{2}CO_{3}$ weighed $1.0235\ g$ . The dissolved mixture was reacted with excess of $Ba(OH)_{2}$ to form $2.1028\ g\ BaCO_{3}$ by the reactions;

$Na_{2}CO_{3}+Ba(OH)_{2}\rightarrow BaCO_{3}+2NaOH$

$NaHCO_{3}+Ba(OH)_{2}\rightarrow BaCO_{3}+NaOH+H_{2}O$

What is the percentage of $NaHCO_{3}$ in the mixture? $[Ba=137]$

1962406_1730875_ans_653613dcd77b4a65a6810086e3358517.jpg

A $2.0\ g$ of sample containing $Na_{2}CO_{3}$ and $NaHCO_{3}$ loses $0.248\ g$ when heated to $300^{o}C$ . What is the percentage of $Na_{2}CO_{3}$ in the mixture.

Write your answer to the nearest integer.

1725760_1730676_ans_69d30a1e74e345e69782022d758745d9.jpg

A precipitate of $AgCl$ and $AgBr$ weighs $0.4066\,g.$ On heating in a current of chlorine, the $AgBr$ is converted to $AgCl$ and the mixture loses $0.0725\,g$ in weight. Find the $\%$ of $Cl$ in origijnal mixture.

1764474_1742789_ans_dd485719eb1e40a9aed76ddcf10b3f7c.jpg

The total pressure of a mixture of $H_2$ and $O_2$ is $1.0$ bar. The mixture is allowed to react to form water, which is completely removed to leave only pure $H_2$ at a pressure of $0.35$ bar. Assuming ideal gas behaviour, and that all pressure measurements were made under the same temperature and volume conditions, calculate the composition of the original mixture.

$Avagadro's Law$ says:

Equal volume of two gases at same pressure and temperature conditions have equal number of molecules.

∴1

$O_2$ molecule consumes 2

$H_2$ molecules.

∴Since after the reaction only

$H_2$ is left, we know all of

$O_2$ is consumed.

∴This implies what all is consumed 1/3 of it was

$O_2$ .

Hence,

$O_2$ in the mix was at a partial pressure of (1 - 0.35)* 1/3 = 0.22

And rest was

$H_2$ = (1 - 0.22) = 0.78

Hence oxygen- hydrogen composition in original mix was:

0.22 bar

$O_2$ and 0.78 bar

$H_2$ .

A $5.0\,g$ sample of a natural gas consisting of $CH_4, C_2H_4$ was burnt in excess of oxygen yielding $14.5\,g\,CO_2$ and some $H_2O$ as product. What is weight percentage of $CH_4$ and $C_2H_4$ in mixture.

2033339_1742758_ans_db751105e2aa4e87adfa41ded748a839.jpg

If the value of Avogadro number is $6.023 \times 10^{23} mol^{-1}$ and the value of Boltzmann constant is $1.380 \times 10^{-23} JK^{-1}$ , then the number of significant digits in the calculated value of the universal gas constant is:

1610048_1733413_ans_a864e9fcc32546cf83d15c6f3b672da1.png

A mixture containing $2.24$ litres of $H_2$ and $1.12$ litres of $D_2$ at NTP is taken inside a bulb connected to another bulb by a stopcock with a small opening. The second bulb is fully evacuated, the stopcock opened for a certain time and then closed. The first bulb is now found to contain $0.10\ g\ H_2$ . Determine the percentage composition by weight of the gases in the second bulb.

1694236_1737092_ans_5bfdef114a984d5585cb2654745fa3c3.jpeg

Fill in the following blank with a suitable word.
Chemical equations are balanced to satisfy the law of ___________.
Enter 1 for 'conservation of mass' or 2 for 'chemical combination'.

Every chemical equation adheres to the law of conservation of mass, which states that matter can neither be created nor destroyed. Therefore, there must be the same number of atoms on both sides of the chemical equation.

When an equal number of atoms of an element is present on both sides of a chemical equation, the equation is said to be balanced.

Hence, the correct answer is the law of conservation of mass.

A sample of Mg metal containing some MgO as impurity was dissolved in $125\,mL$ of $0.1\,N\,H_2SO_4.$ The volume of $H_2$ evolved at $27.5^\circ C$ and 1 atm was $120.0\,mL.$ The resulting solution was found to be $0.02N$ with respect to $H_2SO_4.$ Calculate the weight of sample dissolved and the $\%$ by weight of pure Mg metal in sample. Neglect any charge in volume by weight of pure Mg metal in sample. Neglect any change in volume.

1760403_1742855_ans_1c8f1634814d46ccbf04128b6ed17e6c.jpg

A plant virus is found to consist of uniform cylindrical particles of $150\overset{o}{A}$ in diameter and $5000\overset{o}{A}$ long. The specific volume of the virus is $0.75$ $cm^3/g$ . If the virus is considered to be a single particle, find its molecular weight.

1781296_1740709_ans_342886f01a914a4891c5fcbcebc53437.JPG

Use the following data to calculate Avagodro's number(N). Density of NaCl=2.165 g $cm^{-3}$ . Distance between $Na^{+}$ and $Cl^{-}$ in NaCl=281 pm.

[If $x \times 10^{23}$ is the answer then find the value of $x$ to the nearest integer]

1767728_1743778_ans_4ac0b962aaa14d2aba9d6efb49803b8a.jpg

A gas mixture of 3 litre of propane and butane on complete combustion at $25^\circ C$ produced 10 litre of $CO_2.$ Find out the composition of mixture.

1762589_1743444_ans_8b91ab6deef6457abf8b1d0627c23e41.jpg

How much is 1 a.m.u?

1962579_1824025_ans_53b8f5f46533441490bd9ef30a974f8e.jpg

Fill in the blanks :
The quantity of matter in an object is called its ______.

The quantity of matter in an object is called its

$\underline{\textbf{mass}}$ .

Fill in the blanks
In a chemical reaction , the sum of the masses of the reactants and the products remains unchanged . This is called _____ .

In a chemical reaction, the sum of the masses of the reactants and the products remain unchanged. This is called the Law of Conservation of mass.

Fifty millilitre of a mixture of $CO$ and $CH_{4}$ , was exploded with $85 ml$ of $O_{2}$ . The volume of $CO_{2}$ , produced was $50 ml$ , Calculate the percentage composition of the gaseous mixture.

Let the

$volume\ of\ CO\ = x ml$

$Volume\ of\ CH_{4}, = (50-x) ml$

$\underset{x ml}{CO} + \underset{{\dfrac{x}{2}} ml}{\dfrac{1}{2}O_{2}} \rightarrow \underset{x ml}{CO_{2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)$

$\underset{(50-x)}{CH_{4}}+\underset{2(50-x)}{2O_{2}}\rightarrow \underset{(50-x)ml}{CO_{2}} + 2H_{2}O \ \ \ \ \ \ \ \ ...(ii)$

$Total\ volume\ of\ O_{2} = \dfrac{x}{2} + 2(50-x) = 85$

$\therefore x=10ml$

$Volume\ of\ CO = 10 ml$

$Volume\ of\ CH_{4} = (50 — 10) = 40 ml$

$Percentage\ of\ CO = \dfrac{10 \times 100}{50} = 20\%$

$Percentage\ of\ CH_{4} = (100 — 20) = 80\%$

Why atomic masses are the average values?

Most of the elements exist in different isotopes i.e., atoms with different masses, eg.

$Cl$ has two isotopes with mass numbers

$35$ and

$37$ existing in the ratio

$3 : 1$ . Hence, average value is taken.

Gas	Volume (L)	Number of moles
Hydrogen	224	10
Helium	112	5
Oxygen	224	10
Ammonia	56	2.5

Isotope	$\%$ Natural abundance	Molar mass
$^1 H$	$99.985$	$1$
$^2H$	$0.015$	$2$

Some Basic Concepts Of Chemistry - Class 11 Medical Chemistry - Extra Questions

Calculate the concentration of NaOH solution in g/mL, which has the same normality as that of a solution of HCl of concentration 0.04 g/mL.

Silver crystallises in fcc lattice. If edge length of the cell is 4.07\times {10}^{-8}cm4.07\times {10}^{-8}cm and density is 10.5g10.5g {cm}^{3}{cm}^{3}, calculate the atomic mass of silver.

H_2SO_4H_2SO_4 is 98% by weight of solution, M x 10^2M x 10^2. What is the value of M?

Insulin contains 3.43.4% sulpur calculate minimum molecular weight of insulin.

1 atmosphere means how many Pascal?

Which has the highest mass ?(a) 50 g iron (b) 5 moles of { N }_{ 2 }{ N }_{ 2 }(c) 0.1 mol atom of Ag (d) { 10 }^{ 23 }{ 10 }^{ 23 } atoms of carbon

The total mass of reactants and products in a reaction will always be the same. This law is known as:

The atomic mass of carbon is:

16 g of oxygen and 3 g of hydrogen are mixed and kept at 760 mm pressure and 0. The total volume occupied by the mixture will be nearly(A) 22.4 litre(B) 33.6 litres(C) 448 litres(D) 44800 ml

What is hydrogen azide?

What are the conditions for something to be called as matter?

Define Avogadro's law. Taking a suitable example prove that it is not in contradiction with Dalton's Atomic Theory.

The percentage of 33 element are as follows in calcium carbonate (CaCO_{3})(CaCO_{3}) Calcium 40\%40\%, carbon 12\%12\%, Oxygen 48\%48\%. if law of constant proportion is true what weight of these element will be present in 1.5 g1.5 g of another sample of calcium carbonates (CaCO_{3})(CaCO_{3})

An oxide M_2O_3M_2O_3 contains 10.3 % oxygen by weight. Calculate the atomic weight of M.

A solution contains 50 g of sugar in 350 g of water. Calculate the concentration of solution in terms of mass by mass percent of the solution.

Chlorine occurs in nature in two isotope forms with masses 35u and 37u. The percentage of ^{35}\textrm{Cl}^{35}\textrm{Cl} is 75%. Find the average atomic masses of chlorine atoms.

Calculate the mass of 0.125 mol NaOH

What is one amu or one 'u'?

Define atomic mass unit.

Define the atomic mass unit.

Calculate the weight of ammonia gas required for reacting with sulphuric acid to give 78 g of fertilizer ammonium sulphate.

\text{The following questions are about one mole of sulphuric acid}\ [H_{2}SO_{4}]\text{The following questions are about one mole of sulphuric acid}\ [H_{2}SO_{4}].find the number of gram hydrogen atoms in 1 mole of [H_{2}SO_{4}][H_{2}SO_{4}].

During an experiment the students were asked to prepare a 10% solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water while Sarika prepared it by dissolving 10g of sugar in water to make 100g of the solution. Compare the mass % of the two solutions.

Give reason: Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

The unified atomic mass unit is the standard unit that is used for indicating mass on an atomic or molecular scale. If true enter 1, else enter 0

Mass and percentage by weight do not change with ________ (temperature/volume).

The solubility of salt at 30^{\circ}C30^{\circ}C isCalculate the weight (g) of water required to prepare a saturated solution containing 90 g of salt.

If 11 g of oxalic acid is dissolved in 500 mL of solution (density == 1.1 g mL^{-1}^{-1}), what is the mass % of oxalic acid in solution?

The atomic mass of _{11}^{23}\textrm{Na}_{11}^{23}\textrm{Na} in amu is :

The percent composition by mass of the phosphorus in calcium phosphate Ca_3(PO_4)_2Ca_3(PO_4)_2 is xx. Then, the value of xx is :

Boron has two stable isotopes, ^{10}^{10}B(19%) and ^{11}^{11}B(81%). The atomic mass that should appear for boron in the periodic table is ______ .(Enter the nearest integer value)

The aluminum sulphate hydrate [Al_2(SO_4)_3\cdot xH_2O][Al_2(SO_4)_3\cdot xH_2O] contains 8.208.20 percent AlAl by mass. The value of xx, that is, the number of water molecules associated with each Al_2(SO_4)_3Al_2(SO_4)_3 units is ________. (write the ans in the form of x/3)

Al_2(SO_4)_3.xH_2OAl_2(SO_4)_3.xH_2O has 8.208.20% AlAl by mass. The value of x/3x/3 is _________.

Atomic mass unit is also called Dalton(Da). Explain.

Analysis of a metal chloride XCl_3XCl_3 shows that it contains 67.2\%67.2\% ClCl by mass. The molar mass of XX is __________.

Disilane, Si_2H_xSi_2H_x, is analyzed and found to contain 90.28\%90.28\% of silicon by weight, then the value for xx is (Si=28)(Si=28) :

Na_2SO_3\cdot yH_2ONa_2SO_3\cdot yH_2O has 50\%50\% H_2OH_2O by mass. Thus, yy is :

Match \%\% of carbon in List-I with the compound given in List-II.

When 2.4952.495 gg of CuSO_4\cdot xH_2OCuSO_4\cdot xH_2O (molar mass 249.5249.5) is heated, 0.050.05 mole of H_2OH_2O is lost. Thus, xx is:

A solution _______ \%\% by mass of solvent is 16.67\%16.67\% by mass of solution. Write the value in the form of x\times10x\times10, where xx is:

Total number of ions in aluminium phosphate is :

There are_______moles of anion in one mole of sodium sulphate.

If at a particular temperature, the density of 18\ M \ H_2SO_418\ M \ H_2SO_4 is 1.81.8 g cm^{-3}^{-3}. Percentage concentration (%) by mass of solution is :

Matter is neither .......... nor .......... in a chemical reaction.

Calculate the mass percent of OO present in sodium sulphate \displaystyle \left( { Na }_{ 2 }{ SO }_{ 4 } \right)\displaystyle \left( { Na }_{ 2 }{ SO }_{ 4 } \right).

In diammonium phosphate, (NH_4)_2HPO_4(NH_4)_2HPO_4, the percentage of P_2O_5P_2O_5 is

Describe law of conservation of matter.

The mass of 94.5 ml of a gas at S.T.P. is found to be 0.2231 g. Calculate its molecular mass.

16 grams of NaOHNaOH is dissolved in 100 grams of water at 25^o^oC to form a saturated solution. Find the solubility and weight percentage of the solution.

Compounds are formed by chemically combining elements in a .................... proportion by weight.

Why do atomic masses of most of the elements in atomic mass units involve fraction?

If 10 l10 l of oxygen contains x molecules, then 10 l10 l of ozone contains ______ molecules at the same temperature and pressure.

Calculate the mass of urea NH_2CONH_2NH_2CONH_2 required in making 2.5 kg of 0.25 molal aqueous solution.

Fill in the blanks with suitable words:The atomic mass of an atom is 2323 and its atomic number is 1111. The atom has ________ neutrons.

Fill in the blanks with suitable words.Average atomic mass of chlorine is ____________.

In the brown ring complex, the oxidation state of iron is:

State Avogadro's law?

a) State any three findings of modem atomic theory.b) Write down any two applications of Avogadro's law.

Find the concentration of solution in terms of weight percent if 2020 grams of common salt is dissolved in 5050 grams of water.

Find the concentration of solution in terms of weight percent if 20g20g of common salt is dissolved in 50g50g of water.

Calculate the weight of lime (CaOCaO) that can be obtained by heating 200 kg of limestone which is 93% pure. How many moles of impure potassium chlorate of 75% purity is required to produce 48 g of oxygen?

The relative density of a mixture of nitrogen and oxygen is 14.4 (H = 1) and the relative densities of nitrogen and oxygen are 14.0 and 16.0 (H =1)respectively. Calculate the composition of the mixture (i) by volume and (ii) by mass.

2 g mixture of glucose and sucrose is dissolved in 1 litre water at 298 K to develop an osmotic pressure of 0.207 atm. Calculate percentage composition of glucose and sucrose by mole as well as by mass.

Find the amount of zinc in gram necessary to produce 224ml224ml of {H}_{2}{H}_{2} by the reaction with sulphuric acid.

A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen.Calculate the percentage composition of the compound by weight.

Calculate the weight of lime (CaOCaO) obtained by heating 200kg200kg of 95% pure lime stone (Ca{CO}_{3}Ca{CO}_{3}).Ca{CO}_{3}\rightarrow CaO+{CO}_{2}Ca{CO}_{3}\rightarrow CaO+{CO}_{2}

The self ionisation constant for HCOOHHCOOH is {10}%{-6}{10}%{-6} what percentage of HCOOHHCOOH are converted to HCOOHCOO ion. The density of HCOOHHCOOH is 1.22g1.22g {mL}^{-1}{mL}^{-1}.

Lithium exists in nature in the form of two isotopes, Li-6 and Li-7 with atomic masses 6.0151u and 7.0160u and the percentages 8.24 and 91.76 respectively. Calculate average atomic mass.

State the law of conservation of mass. State the main points of Landolt's experiment for experimental evidence of the law.

Discuss about determining mass percentage of oil from oil seed.

Calculate the percentage loss of mass of hydrated copper[II] sulphate [CuSO_4.5H_2O][CuSO_4.5H_2O] when it is completely dehydrated. CuSO_4.5H_2O \to CuSO_4+5H_2OCuSO_4.5H_2O \to CuSO_4+5H_2O[Atomic weights are Cu = 64, S=32, O=16, H=1Cu = 64, S=32, O=16, H=1]

92 g mixture of CaCO_3CaCO_3 and MgCO_3MgCO_3 heated strongly in an open vessel. After complete decomposition of the carbonates it was found that the weight of residue left behind is 48 g. Find the mass of MgCO_3MgCO_3 in grams in the mixture.

Calculate the mass per cent of different elements present in sodium sulphate (Na_2SO_4)(Na_2SO_4).

Claculate the mass % of different element in Na_2SO_4Na_2SO_4 ?

In a Carius determination, 0.234g0.234g of an organic substance gave 0.334g0.334g of barium sulphate. Calculate the percentage of sulphur in the given compound. [Ba = 137, S=32, O=16][Ba = 137, S=32, O=16]

Calculate the no. of atoms in:a) 0.5 mol atom of nitrogenb) 0.2 mol molecules of hydrogenc)3.2 g of sulphur (Z for sulphur=32)

State and explain law of conservation of mass.

A solution contains 40 grams of common salt dissolved in 345 grams the solution. Calculate the concentration percentage of the solution by mass percentage.

0.246 g of an organic compound on complete combustion gave 0.198 g of carbondioxide and 0.1014 g of water, then the percentage composition of carbon and hydrogen in the compound respectively1) 4.58, 21.95 2) 21.95, 4.583) 45.8, 2.195 4) 2.195, 45.8

Silver crystallises in fcc lattice. If edge length of the cell is $4.07\times {10}^{-8}cm$ and density is $10.5g$ ${cm}^{3}$ , calculate the atomic mass of silver.

$H_2SO_4$ is 98% by weight of solution, $M x 10^2$ . What is the value of M?

Insulin contains $3.4$ % sulpur calculate minimum molecular weight of insulin.

Which has the highest mass ?
(a) 50 g iron (b) 5 moles of ${ N }_{ 2 }$
(c) 0.1 mol atom of Ag (d) ${ 10 }^{ 23 }$ atoms of carbon

16 g of oxygen and 3 g of hydrogen are mixed and kept at 760 mm pressure and 0. The total volume occupied by the mixture will be nearly
(A) 22.4 litre
(B) 33.6 litres
(C) 448 litres
(D) 44800 ml

The percentage of $3$ element are as follows in calcium carbonate $(CaCO_{3})$ Calcium $40\%$ , carbon $12\%$ , Oxygen $48\%$ . if law of constant proportion is true what weight of these element will be present in $1.5 g$ of another sample of calcium carbonates $(CaCO_{3})$

An oxide $M_2O_3$ contains 10.3 % oxygen by weight. Calculate the atomic weight of M.

Chlorine occurs in nature in two isotope forms with masses 35u and 37u. The percentage of $^{35}\textrm{Cl}$ is 75%. Find the average atomic masses of chlorine atoms.

$\text{The following questions are about one mole of sulphuric acid}\ [H_{2}SO_{4}]$ .
find the number of gram hydrogen atoms in 1 mole of $[H_{2}SO_{4}]$ .

The solubility of salt at $30^{\circ}C$ isCalculate the weight (g) of water required to prepare a saturated solution containing 90 g of salt.

If 11 g of oxalic acid is dissolved in 500 mL of solution (density $=$ 1.1 g mL $^{-1}$ ), what is the mass % of oxalic acid in solution?

The atomic mass of $_{11}^{23}\textrm{Na}$ in amu is :

The percent composition by mass of the phosphorus in calcium phosphate $Ca_3(PO_4)_2$ is $x$ . Then, the value of $x$ is :

Boron has two stable isotopes, $^{10}$ B(19%) and $^{11}$ B(81%). The atomic mass that should appear for boron in the periodic table is ______ .
(Enter the nearest integer value)

The aluminum sulphate hydrate $[Al_2(SO_4)_3\cdot xH_2O]$ contains $8.20$ percent $Al$ by mass. The value of $x$ , that is, the number of water molecules associated with each $Al_2(SO_4)_3$ units is ________. (write the ans in the form of x/3)

$Al_2(SO_4)_3.xH_2O$ has $8.20$ % $Al$ by mass. The value of $x/3$ is _________.

Analysis of a metal chloride $XCl_3$ shows that it contains $67.2\%$ $Cl$ by mass. The molar mass of $X$ is __________.

Disilane, $Si_2H_x$ , is analyzed and found to contain $90.28\%$ of silicon by weight, then the value for $x$ is $(Si=28)$ :

$Na_2SO_3\cdot yH_2O$ has $50\%$ $H_2O$ by mass. Thus, $y$ is :

Match $\%$ of carbon in List-I with the compound given in List-II.

When $2.495$ $g$ of $CuSO_4\cdot xH_2O$ (molar mass $249.5$ ) is heated, $0.05$ mole of $H_2O$ is lost. Thus, $x$ is:

A solution _______ $\%$ by mass of solvent is $16.67\%$ by mass of solution. Write the value in the form of $x\times10$ , where $x$ is:

If at a particular temperature, the density of $18\ M \ H_2SO_4$ is $1.8$ g cm $^{-3}$ . Percentage concentration (%) by mass of solution is :

Calculate the mass percent of $O$ present in sodium sulphate $\displaystyle \left( { Na }_{ 2 }{ SO }_{ 4 } \right)$ .

In diammonium phosphate, $(NH_4)_2HPO_4$ , the percentage of $P_2O_5$ is

16 grams of $NaOH$ is dissolved in 100 grams of water at 25 $^o$ C to form a saturated solution. Find the solubility and weight percentage of the solution.

If $10 l$ of oxygen contains x molecules, then $10 l$ of ozone contains ______ molecules at the same temperature and pressure.

Calculate the mass of urea $NH_2CONH_2$ required in making 2.5 kg of 0.25 molal aqueous solution.

Fill in the blanks with suitable words:
The atomic mass of an atom is $23$ and its atomic number is $11$ . The atom has ________ neutrons.

Fill in the blanks with suitable words.
Average atomic mass of chlorine is ____________.

a) State any three findings of modem atomic theory.
b) Write down any two applications of Avogadro's law.

Find the concentration of solution in terms of weight percent if $20$ grams of common salt is dissolved in $50$ grams of water.

Find the concentration of solution in terms of weight percent if $20g$ of common salt is dissolved in $50g$ of water.

Calculate the weight of lime ( $CaO$ ) that can be obtained by heating 200 kg of limestone which is 93% pure. How many moles of impure potassium chlorate of 75% purity is required to produce 48 g of oxygen?

Find the amount of zinc in gram necessary to produce $224ml$ of ${H}_{2}$ by the reaction with sulphuric acid.

Calculate the weight of lime ( $CaO$ ) obtained by heating $200kg$ of 95% pure lime stone ( $Ca{CO}_{3}$ ).

$Ca{CO}_{3}\rightarrow CaO+{CO}_{2}$

The self ionisation constant for $HCOOH$ is ${10}%{-6}$ what percentage of $HCOOH$ are converted to $HCOO$ ion. The density of $HCOOH$ is $1.22g$ ${mL}^{-1}$ .

Calculate the percentage loss of mass of hydrated copper[II] sulphate $[CuSO_4.5H_2O]$ when it is completely dehydrated.
$CuSO_4.5H_2O \to CuSO_4+5H_2O$
[Atomic weights are $Cu = 64, S=32, O=16, H=1$ ]

92 g mixture of $CaCO_3$ and $MgCO_3$ heated strongly in an open vessel. After complete decomposition of the carbonates it was found that the weight of residue left behind is 48 g. Find the mass of $MgCO_3$ in grams in the mixture.

Calculate the mass per cent of different elements present in sodium sulphate $(Na_2SO_4)$ .

Claculate the mass % of different element in $Na_2SO_4$ ?

In a Carius determination, $0.234g$ of an organic substance gave $0.334g$ of barium sulphate. Calculate the percentage of sulphur in the given compound. $[Ba = 137, S=32, O=16]$

Calculate the no. of atoms in:
a) 0.5 mol atom of nitrogen
b) 0.2 mol molecules of hydrogen
c)3.2 g of sulphur (Z for sulphur=32)

0.246 g of an organic compound on complete combustion gave 0.198 g of carbondioxide and 0.1014 g of water, then the percentage composition of carbon and hydrogen in the compound respectively
1) 4.58, 21.95 2) 21.95, 4.58
3) 45.8, 2.195 4) 2.195, 45.8

If atomic eq. 40% of $PCl_{ 5 }$ dissociated. Calculate equivalent molar mass.

Calculate the mass percentage composition of the elements in nitric acid : $(H=1;N=14,O=16)$

Calculate the total number of electrons present in $18\ mL$ of water.

How much amount of $CaCO_3$ in gram having percentage purity 50 per cent produces 0.56 litre of $CO_2$ at on heating?

Lithium exists in nature in the form of two isotopes, $Li-6$ and $Li-7$ with atomic masses $6.0151u$ and $7.0160u$ and the percentage $8.24$ and $91.76$ respectively.
Calculate average atomic mass.

The half life of the nuclide $Rn^{220}$ is $54.5$ sec. Find mass (in kg) of radon is equivalent to $1$ millicurie?

Calculate the amount of glucose required to prepare $250$ g of $5\%$ solution of glucose by mass.

$4.6 cm^3$ of methyl alcohol is dissolved in $25.2$ g of water. Calculate $\%$ by mass of methyl alcohol.
(Given density of methyl alcohol = $0.7952 gcm^{-3}$ and $C = 12, H = 1, O = 16$ )

What is the mass of the precipitate formed when $50$ mL of $16.9\%$ (w/v) solution of $AgNO_3$ is mixed with $50$ mL of $5.8\%$ (w/v) $NaCl$ solution?
[ $Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5$ ]

An atom bomb is weighing 1 kg explodes releasing $9 \times 10^{13}$ joule of energy. What percentage of mass is converted into energy?

Calculate the percentage composition in terms of mass of a
solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.

Phosphoric
acid is widely used in carbonated beverages , detergents , toothpastes and
fertilizers . Calculate the mass percentage of H , P AND O in phosphoric acid
if atomis mases are H =1 , P=31 and O = 16

In a gaseous mixture 2mol of ${ CO }_{ 2 }$ , 1 mol of ${ H }_{ 2 }$ and 2 mol of He are present than determine mole percentage of ${ CO }_{ 2 }$ .

At $-50^0C$ , liquid $NH_3$ has ionic product is $10^{-30}$ . How many amide $(NH_2^-)$ ions are present per $mm^3$ in pure liquid $NH_3$ ? (take $N_A = 6 \times 10^{23}$ )

Calculate the mass of potassium chlorate required to liberate 6.72 of oxygen at STP.
[Assume: molar mass of potassium chlorate as 122.5 g / mol].

44.8 L of $CO_{2}$ at NTP is obtained by heating $x$ g of pure $CaCO_{3}.x$ is:

By the reaction of carbon and oxygen, a mixture of $CO$ and $CO_{2}$ is obtained. What is the composition(% by mass) of the mixture obtained when 20 grams of $O_{2}$ reacts with 12 grams of carbon ?

A solution of ethanol in water is $10%$ by volume. If the solution and pure ethanol have densities of $0.9866\ g/cc$ and $0.785\ g/cc$ respectively, find the percent by weight.

If the atomic mass of sodium is $23u$ , what will be the atomic mass of sodium ion?

$15.9 \,g$ of copper sulphate and $10.6 \,g$ of sodium carbonate react together to give $14.2 \,g$ of sodium sulphate and $12.3\,g$ of copper carbonate. Which law of chemical combination is obeyed? How?

A solution of ethanol in water is $10%$ by volume. If the solution and pure ethanol have densities of $0.9866 glcc$ and $0.785 g/cc$ respectively, find the percent by weight.

The mole fraction of benzene in a solution in toluene is $0.40$ . Calculate the weight per cent of benzene in the solution.

$27.6$ g ${K_2}C{O_3}$ was treated by a series of reagents so as to convert all of its carbon to ${K_2}Z{n_3}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]_2}$ The weight of product is?

The percent loss in weight after heating a pure sample of potassium chlorate (mol. wt. = 122.5) will be ( ${KCIO }_{3 }\rightarrow KCI +{ O}_{2 }$ )

At STP, 2.8 liters of hydrogen sulphide were mixed with 1.6 liters of sulphur dioxide and reaction occurred according to the equation :
${ 2H }_{ 2 }S(g)+{ SO }_{ 2 }(g)\longrightarrow { 2H }_{ 2 }O(l)+3S(s)$

Calculate the molality in the aqueous solution containing $3.0gm$ of urea in $250gm$ of water
( $H=1,N=14,O=16$ )

Evaporation of water occurs at all temperature
How is eveporation dependent on temperature ?

Determine the percentage composition of each atom in the molecule of ${ Na } _ { 2 } { CO } _ { 3 }$ .

The weight of 350 mL of a diatomic gas at $0^0C$ and 2 atm pressure is 1 g. The wt. of one atom is

a) Define atomic mass unit.
b) Distinguish between molecular mass and molar mass.
c) Give an example of (i) diatomic, (ii) triatomic molecule of compounds.

How many grans of concentrated sitric acid $sol^n$ should be used to prepare $250$ mL of $2$ M $HNO_3$ ?
The concentrated acid is $70\%$ $HNO_3$ ?

An atom of an element is $10.1$ times heavier than the mass of a carbon atom. What is its mass in a.m.u?

A gaseous mixture of $H_{2}$ and $N_{2}O$ gas contains 66 mass $\%$ of $N_{2}O$ . What is the average molecular mass of mixture?

How many amu's are present in ${10^{ - 3}}$ g?

1 L of $CO_2$ is passed over hot coke.When the volume of reaction mixture becomes 1.4 L, the composition of reaction mixture is

The saturation temperature for 20.7g of $CuSO_4$ soluble in water is_______.