MCQ Questions for Class 12 Engineering Chemistry Solutions Quiz 4

When

$1$ mole of a substance is present in

$1$ L of the solution, it is known as :

0%
normal solution
0%
molar solution
0%
molal solution
0%
None of the above

Explanation

Molar concentration is a measure of the concentration of a solute in a solution and its unit is mol L

$^{-1}$ . Molarity is a method to express the concentration of a solution. It is defined as the number of moles of solute dissolved per liter of solution.

Hence, when

$1$ mole of a substance is present in

$1$ L of the solution, it is known as a molar solution.

Option B is correct.

Van't Hoff factor of

$0.01M$

$Ba{Cl}_{2}$ is

$1.98$ , percentage dissociation of

$Ba{Cl}_{2}$ on this concentration will be:

0%
$69$
0%
$100$
0%
$49$
0%
$98$

Explanation

$Ba{Cl}_{2}\rightleftharpoons {Ba}^{2+}+2{Cl}^{-}$

Initially

$1$

$0$

After disso.

$1-\alpha$

$\alpha$

$2\alpha$

Total moles after dissociation

$=1-\alpha+\alpha+2\alpha$

$=1+2\alpha$

$i=\cfrac { observed\quad moles\quad of\quad solute }{ normal\quad moles\quad of\quad solute } =\cfrac { 1+2\alpha }{ 1 }$

$\therefore \quad \alpha =\cfrac { i-1 }{ 2 } =\cfrac { 1.98-1 }{ 2 } =\cfrac { 0.98 }{ 2 } =0.49$

For

$1$ mole

$\alpha=0.49$

For

$0.01$ mole

$\alpha=\cfrac{0.49}{0.01}=49$ %

Hence, the correct option is

$C$

What is the effect of temperature on solubility of a gas?

0%
Solubility of a gas increase with rise in temperature
0%
Solubility of a gas decreases with rise in temperature
0%
No effect
0%
Sometimes increases with increase in temperature, sometimes decreases

Explanation

The solubility of a gas decreases with rising temperatures. According to Charle's law, as we increase temperature, the volume of a given mass of gas dissolved in solution increases.

Hence, water cannot accommodate gas in it and the gas bubbles out.

So, the correct option is

$B$

If solubility is not changing as the temperature increases then which graph is solubility curve of the substance?

The following substances were all dissolved in 100 grams of water at 290 K to produce saturated solutions.If the solution is heated to 310 K, which substance will have a decrease in its solubility:

0%
$NaCI$
0%
$KI$
0%
$CaCl_2$
0%
$HCI$
0%
$KNO_3$

Which of the following is not miscible in alcohol?

0%
Water
0%
Petrol
0%
Diesel
0%
Kerosene

The solubility of a solute indicates :

0%
the temperature of the solution
0%
the quantity of solvent
0%
the quantity of solute
0%
the nature of the solute and solvent
0%
all of these

At equilibrium, the rate of dissolution of a solid solute in a volatile liquid solvent is:

0%
less than the rate of crystallisation
0%
greater than the rate of crystallisation
0%
equal to the rate of crytallisation
0%
zero

A closed vessel is maintained at a constant temperature. It is first evacuated and then vapour is injected into it continuously. The pressure of the vapour in the vessel is:

0%
increases continuously
0%
first increases and then remains constant
0%
first increases and then decreases
0%
none of the above

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Hydrogen gas adsorbed on surface of platinum is an example of solid in gas solution.

0%
True
0%
False

Which of the following is a possible source of error in the results of this experiment?

0%
Inaccurate weighing of the isopropanol
0%
The presence of bits of broken glass in the flask
0%
Failure to correct for humidity
0%
Misreading of the atmospheric pressure

The initial pressure inside the flask is:

0%
near zero (vacuum)
0%
$27\ mm\$ $\displaystyle Hg$
0%
$760\ mm\$ $\displaystyle Hg$
0%

$(760-27)\ mm\ Hg$

Explanation

Since here in the figure given the mercury is at the same height in both the arms containing the sample and the arm is atom to the atmosphere, the pressure inside the flask initially is

$760mHg$ .

Option (C) is correct.

Which of the following salt has the same value of Van't Hoff factor i as that of

$K_3 [Fe(CN)_6]$ ?

0%
$Al_2(SO_4)_3$
0%
$NaCl$
0%
$Na_2SO_4$
0%
$Al(NO_3)_3$

Explanation

Van't Hoff factor of

$K_3[Fe(CN)_6]$

$K_3[Fe(CN)_6]\rightarrow 3K^+ + [Fe(CN)_6]^-$

$i=4$

$i_{Al_2(SO_4)_3}=5$

$i_{NaCl}=2$

$i_{Na_{SO_4}}=3$

$i_{Al_2(NO_3)_3}=4$

thus, the correct answer is D.

Which of the following is a true solution?

0%
$NaCl$ is sulphur dioxide
0%
Copper in silver
0%
Salt in petrol
0%
Mud in water

Explanation

Hint: Homogeneous mixture of two or more substances is called a true solution.

Correct Answer: Option B

Explanation:

$NaCl$ in sulfur dioxide $SO_2$ forms molten sodium sulfate. So, it is not a true solution.

$Cu$ and $Ag$ melts together to form an alloy. So, it is a true solution.

Petrol is a non-polar solvent. So, salt ( $NaCl$ ) is insoluble in petrol.

Mud in water forms a colloidal solution.

Hence, copper in silver is a true solution.

Identify which of the following statements is FALSE?

0%
The vapor pressure of a liquid decreases with increasing atmospheric pressure.
0%
The value of an equilibrium constant is dependent on temperature.
0%
The rate of a spontaneous reaction cannot be determined solely by its Gibbs free energy.
0%
During a phase transition, the temperature of a substance must be constant.
0%
The addition of a catalyst to a reaction at equilibrium has no net effect on the system.

In which of the following solvent, the solubility of

$NaCl$ is maximum ?

0%
Methanol
0%
Water
0%
Formic acid
0%
Formamide

Explanation

$NaCl$ is an ionic compound. So it's solubility is maximum in another ionic compound. So water is the only ionic compound in the given options, so B is the correct answer.

The mass of a non-volatile solute of molar mass 40 g

$mol^{-1}$ that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is :

0%
10 g
0%
11.4 g
0%
9.8 g
0%
12.8 g

Explanation

The mass of a non-volatile solute of molar mass 40 g /mol that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is 10 g.

Let w g of the non-volatile solute is used. Its molar mass is 40 g/mol.

The number of moles of non volatile solute

$\displaystyle = \dfrac {w}{40} =0.025w$ moles.

114 g of octane (molar mass 114 g/mol) corresponds to

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

The mole fraction of non volatile solute is

$\displaystyle = \dfrac {0.025w}{1+0.025w}$ . The mole fraction of non volatile solute is proportional to the relative lowering in the vapour pressure.

$\displaystyle \dfrac {P^0-P}{P^0}=X_B$

$\displaystyle \dfrac {20}{100}= \dfrac {0.025w}{1+0.025\ w}$

$\displaystyle 100 (0.025w)= 20(1+0.025w)$

$\displaystyle 2.5w=20+0.50w$

$\displaystyle 2w=20$

$\displaystyle w=10$ g

Hence, the correct option is

$\text{A}$

Which of the following equation does NOT correctly depict the process by which the given solute dissolves in water?

0%
Solid calcium nitrate is added to water to form an aqueous solution.

$Ca(NO_{3})_{2}(s) \rightarrow Ca^{2+} + 2NO_{3}$ $^{-}(aq)$
0%
Magnesium hydroxide, a very slightly soluble solid, is dissolved in water.

$Mg(OH)_{2}(s)\rightarrow Mg^{2+}(aq) + 2OH^{-}(aq)$
0%
Sucrose, table sugar, $C_{12}H_{22}O_{11}$ , dissolves in water.

$C_{12}H_{22}O_{11}(s) \rightarrow C_{12}H_{22}O_{11}(aq)$
0%
Phosphoric acid, a weak triprotic acid, dissolves in water.

$H_{3}PO_{4}(l)\rightarrow 3H^{+}(aq) + PO_{4}^{3-}(aq)$

Explanation

All the options are correct

Since

$H_3PO_4$ is a tricrotic acid but its degree of dissociation is not 100%, so it does not give

$3H^+$ ion on adding water to it.
Hence option D not correctly depict the process

Which of these is a colligative property of a solution?

0%
Relative lowering of vapour pressure
0%
Elevation of boiling point
0%
Depression of freezing point
0%
Osmotic pressure

Which of the following solutions does not show positive deviation?

0%
Acetone and Ethanol
0%
Toluene and Benzene
0%
Water and Ethanol
0%
Water and Methanol

Explanation

$\begin{array}{l}\text { Toluene and Benzene are very similar molecules } \\\text { and hence are ideal solution, } \\\text { others are solutions showing positive deviation } \\\text { (Theory based. To be memorized.) } \\\text { Hence, } B \text { is correct. }\end{array}$

The van't Hoff factor of acetic acid solution is __________ than one and the value of normal colligative property is __________ than the observed colligative property of this solution.

0%
more, less
0%
less, more
0%
less, less
0%
more, more

Explanation

The Van't Hoff factor of acetic acid solution is more than one and the value of normal colligative property is less than the observed colligative property of this solution. Acetic acid dissociates into acetate ions and hydrogen ions.

The Van't Hoff factor

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

The Van't Hoff factor

$\displaystyle i = \dfrac {n\text {(observed)}}{ \text {n(theoretical)}}$

The Van't Hoff factor

$\displaystyle i = \dfrac {\text {observed colligative property of electrolyte solution}}{ \text {observed colligative property of non electrolyte solution}}$

Volume of mixing for a liquid solution is :

0%
may be positive or negative depending on the nature of liquids
0%
0
0%
+ve
0%
-ve

Explanation

$\begin{array}{l}\text { Depending on the nature of two solution }\\\Delta V=0 \text { for ideal solutions } \\\Delta V>0 \text { for solutions showing positive deviation } \\\Delta V \text { < } 0 \text { for solutions showing negative deviation } \\\text { Hence, } A \text { is correct. }\end{array}$

Which of the following sentences is true?

0%
Relative lowering of vapor pressure of an ideal solution containing a non-volatile solute is equal to the mole fraction of the solute at a given temperature
0%
Relative lowering of vapor pressure of an ideal solution containing a non-volatile solute is equal to the mole fraction of the solvent at a given temperature
0%
Relative lowering of vapor pressure of an ideal solution containing a non-volatile solute depends on the nature of the solute at a given temperature
0%
Relative lowering of vapor pressure of an ideal solution containing a non-volatile solute depends on the nature of the solvent at a given temperature

Explanation

Relative lowering of vapor pressure of an ideal solution containing a non-volatile solute is equal to the mole fraction of the solute at a given temperature.
If

$p^o_A$ is the vapor pressure of pure solvent. And

$p_A$ is the vapor pressure of the ideal solution after a non-volatile solute is added.
The relative lowering of vapor pressure of the solution becomes

$\dfrac{p^o_A-p_A}{p^o_A}$

$=$

$X_B$ , where

$X_B$ is the mole fraction of solute.

$3$ gms of urea is dissolved in

$45$ gms of

${H}_{2}O$ . The relative lowering in vapour pressure is :

0%
$0.05$
0%
$0.04$
0%
$0.02$
0%
$0.01$

Explanation

Relative lowering of vapour pressure is given by:

$\dfrac{P^o-P}{P^o}=X_{solute}=\dfrac{n_2}{n_1}=\dfrac{\dfrac{3}{60}}{\dfrac{45}{18}}=0.02$

In an endothermic process, solubility increases with ______ in temperature.

0%
increase
0%
decrease
0%
remains same
0%
none of the above

The correct value for the relative lowering of vapour pressure of a very dilute solution is equal to:

0%
mole fraction of solute
0%
mole fraction of solvent
0%
ratio of moles of solute to moles of solvent
0%
ratio of moles of solvent to moles of solute

Explanation

we have,

$\dfrac{P^0_B - P_t}{P^0_B} = x_A$

or we have

$\dfrac{P^0_B - P_t}{P^0_B} = x_A$ =

$\dfrac{n_A}{n_A + n_B}$

the value of

$n_A$ is very samall for dilute solution, so we have;

$\dfrac{P^0_B - P_t}{P^0_B} = x_A$ =

$\dfrac{n_A}{ n_B}$

When heating of a miscible solution begins, vapours formed will be:

0%
of liquid lower in boiling point
0%
of liquid higher in boiling point
0%
of both liquids with a higher concentration of liquid having low boiling point
0%
None of the above

Which of the following will change the relative lowering of vapour pressure of a solution?

0%
Chemical properties of solute
0%
Amount of solvent
0%
Amount of solute
0%
All of these

Relative lowering of vapour pressure of a solution is equal to:

0%
the mole fraction of solvent
0%
the ratio of moles of solute to moles of solvent
0%
the mole fraction of solute
0%
the ratio of moles of solvent to moles of solute

The equation that represents van't Hoff general solution equation is :

0%
$\pi = \dfrac{n}{V}RT$
0%
$\pi = nRT$
0%
$\pi = \dfrac{V}{n}RT$
0%
$\pi = nVRT$

Explanation

Van't Hoff general equation is

$\pi =\dfrac { n }{ V } RT$

Where,

$\pi$ is pressure,

$T$ is temperature,

$R$ is gas constant,

$n$ is number of moles and

$V$ is volume.

A student slowly mixes salt into 25 ml of water until no more salt dissolves in it. The student could make more salt dissolve in the solution by:

0%
cooling it
0%
evaporating it
0%
heating it
0%
stirring it

$283 K$ a saturated solution of solid

$X$ can be prepared by dissolving

$21.0 g$ of it in

$100 g$ of water. The maximum amount of

$X$ which can be dissolved in

$100 g$ of water at

$313 K$ is

$62.0 g$ . An attempt is made to dissolve

$50.0 g$ of

$X$ in

$100 g$ of water at

$313 K$ .

A. All the

$50.0 g$ of

$X$ will dissolve at

$313 K$
B. At

$313 K$

$29.0 g$ of

$X$ will remain undissolved
C. Solubility of

$X$ decreases with increases of temperature
D. On cooling the solution of

$X$ from

$313 K$ to

$283 K$ more than

$21.0 g$ of

$X$ will crystallize out.

Which of the above statements are correct?

0%
A and B
0%
A and D
0%
B and C
0%
A, C and D

Assuming ideal behaviour, the enthalpy and volume of mixing of two liquids, respectively, are:

0%
zero and zero
0%
+ve and zero
0%
-ve and zero
0%
-ve and -ve

Explanation

Assuming ideal behaviour,

$\Delta H_{mix}=0$ and

$\Delta V_{mix}=0$ respectively.

Option A is correct.

The value of van't Hoff factor for

$0.1$ M

$Ba(NO_3)_2$ solution is

$2.74$ . The degree of dissociation is:

0%
$100\%$
0%
$92\%$
0%
$87\%$
0%
$74\%$

Explanation

$Ba(NO_3)_2$ dissociate as:

Initial at equilibrium,

$\underset{\underset{(0.1-x)M}{0.1M}}{Ba(NO_3)_2}\leftrightharpoons \underset{\underset{aM}{x}}{Ba^{2+}_0}+\underset{\underset{2a M}{2x}}{\underset{0}{2NO^-_3}}$ .
Thus, n

$=3$

$i=n\alpha + (1-\alpha)=3\alpha +1-\alpha$ (per mole)
Van't Hoff factor

$i=1+2\alpha \Rightarrow 2.74=1+2\alpha$

$2\alpha =1.74$

$\alpha =0.87$

$\alpha(\%)=87\%$ .

For a non-volatile solute:

0%
vapour pressure of solution is more than vapour pressure of solvent
0%
vapour pressure of solvent is zero
0%
vapour pressure of solute is zero
0%
all of the above

The pair of boiling point and compound are given as,

$C_6 H_6,$

$CH_3OH$

$C_6H_5NO_2$

$C_6H_5NH_2$

$80^oC$

$65^oC$

$212^oC$

$184^oC$
(I) (II) (III) (IV)

Which will show lowest vapour pressure at room temperature?

0%
$C_6H_6$
0%
$CH_3 OH$
0%
$C_6H_5NO_2$
0%
$C_6H_5NH_2$

Explanation

As the boiling point is inversely proportional to vapour pressure hence, nitrobenzene having bp

$212^oC$ will show the lowest vapour pressure at room temperature.

This is because low vapour pressure makes the substance more non-volatile and thus increase its boiling point.

Option C is correct.

Vapour pressure of pure

$A(p^0_A)=100$ mm Hg
Vapour pressure of pure

$B(p^0_B)=150$ mm Hg

2 mole of liquid A and 3 mole of liquid B are mixed to form an ideal solution. The vapour pressure of the solution will be :

0%
135 mm
0%
130 mm
0%
140 mm
0%
145 mm

For which among the following equimolar aqueous solutions Van't Hoff factor has the lowest value?

0%
Aluminium Chloride
0%
Potassium Sulphate
0%
Ammonium Chloride
0%
Urea

Explanation

For the aqueous solution of urea, van't Hoff factor has the lowest value.

Urea is non-electrolyte and does not dissociate in aqueous solution. Hence, Van't Hoff factor is

$1.$

For aluminium chloride

$\displaystyle AlCl_3$ ,

$\displaystyle i = 4$

$\displaystyle AlCl_3 \rightarrow Al^{3+} + 3Cl^-$

For potassium sulphate

$\displaystyle K_2SO_4$ ,

$\displaystyle i = 3$

$\displaystyle K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$

Following properties will decrease with increase in temperature except:

0%
surface tension
0%
viscosity
0%
density
0%
vapour pressure

6gm of urea is dissolved in 90gm of water. Relative lowering of vapour pressure is:

0%
0.02
0%
0.2
0%
0.002
0%
0.04

Explanation

Relative lowering of vapour pressure,

$\displaystyle \dfrac {\Delta P}{P} = \dfrac { \dfrac {W_2}{M_2} }{ \dfrac {W_1}{M_1} }$

Relative lowering of vapour pressure,

$\displaystyle \dfrac {\Delta P}{P} = \dfrac { \dfrac { \text { 6 g } }{ \text { 60 g/mol }} }{ \dfrac { \text { 90 g}}{ \text { 18 g/mol }} }$

Relative lowering of vapour pressure

$\displaystyle \dfrac {\Delta P}{P} = \dfrac { 0.1 }{ 5 }$

Relative lowering of vapour pressure

$\displaystyle = 0.02$

Mole fraction of the component A in vapour phase is

$x_1$ and mole fraction of component A in liquid mixture is

$x_2$ , then (

$p^0_A =$ vapour pressure of pure A;

$p^0_B =$ vapour pressure of pure B), the total vapour pressure of liquid mixture is :

0%
$p^0_A\dfrac{x_2}{x_1}$
0%
$p^0_A\dfrac{x_1}{x_2}$
0%
$p^0_B\dfrac{x_1}{x_2}$
0%
$p^0_B\dfrac{x_2}{x_1}$

Explanation

Mole fraction of component A in vapour phase is

$x_1$

that means

$P_A = P^0_A * x_1$

and mole fraction of compound A in liquid mistureis

$x_2$

that means we have,

$P_A = x_2 * P^0_A$

so, from both equation we get

$x_2 * P^0_A = x_1 * P_t$

so,

$P_t = \dfrac{x_2}{x_1}* P^0_A$

Which of the following will form an ideal solution?

0%
$C_2H_5OH$ and water
0%
$HNO_3$ and water
0%
$CHCl_3$ and $CH_3COCH_3$
0%
$C_6H_6$ and $C_6H_5CH_3$

Explanation

$C_6H_6$ and

$C_6H_5CH_3$ will form an ideal solution over the entire range of composition. The bonding interactions among the molecules will be same before and after mixing.

Hence, option (D) is correct.

The vapour pressure of pure benzene at

$88^oC$ is 957 mm and that of toluene at the same temperature is 379.5 mm. Calculate the composition of benzene-toluene mixture boiling at

$88^oC$ .

0%
$x_{benzene}=0.66; \,x_{toluene}=0.34$
0%
$x_{benzene}=0.34; \,x_{toluene}=0.66$
0%
$x_{benzene} = x_{toluene}=0.5$
0%
$x_{benzene}=0.75;\, x_{toluene}=0.25$

$25^oC$ the vapour pressure of pure benzene is 93.9 torr. When a nonvolatile solute is added to benzene, the vapour pressure of benzene is lowered to 91.5 torr. Calculate the mole fraction of the solute.

0%
0.974
0%
0.026
0%
0.034
0%
0.056

Van't Hoff factor of

$Hg_2Cl_2$ in its aqueous solution will be (

$Hg_2Cl_2$ is

$80\%$ ionized in the solution):

0%
$1.6$
0%
$2.6$
0%
$3.6$
0%
$4.6$

Explanation

$Hg_2Cl_2\rightleftharpoons Hg^{2+}_2 + 2Cl^-$

Here,

$n=3$

Now,

$\alpha = \dfrac{i-1}{n-1}$

$0.8=\dfrac{i-1}{3-1}$

$i=2.6$

$3g$ urea is dissolved in

$45g$ of water. The relative lowering of vapour pressure is:

0%
$0.05$
0%
$0.04$
0%
$0.02$
0%
$0.01$

Explanation

$n_B=\dfrac{3}{60}=0.05; n_A=\dfrac{45}{18}=2.5$

Now,

$\dfrac{\triangle p}{p_0}=x_B=\dfrac{0.05}{2.5+0.05}=0.0196=0.02$

What is the vapour pressure of a solution of glucose which has an osmotic pressure of 3 atmospheres at

$20^C?$ The vapour pressure of water at

$20^oC$ is

$17.39 mm$ . Consider the density of solution equal to that of the solvent.

0%
$12.35 mm$
0%
$14.35 mm$
0%
$16.35 mm$
0%
$17.35 mm$

$25^oC$ , the vapour pressure of pure benzene is 100 torr, while that of pure ethyl alcohol is 44 torr. Assuming ideal behaviour, calculate the vapour pressure at

$25^oC$ of a solution which contains 10 g of each substance.

0%
33.775 torr
0%
54.775 torr
0%
64.775 torr
0%
60.775 torr

Explanation

We have,

moles of benzene =

$\dfrac{10}{78}$ = 0.128

moles of ethyl alcohol =

$\dfrac{10}{46}$ = 0.217

$P_t = P^0_A * x_A + P^0_B * x_B$

$P_t = 100 * \dfrac{0.128}{0.345} + 44*\dfrac{0.217}{.345}$

$P_t = 64.77 torr$

Which of the following is not correct for an ideal solution?

0%
Raoult's law is obeyed for entire concentration range and temperatures
0%
$\triangle H_{mix}=0$
0%
$\triangle V_{mix}=0$
0%
$\triangle S_{mix}=0$

Explanation

For an ideal solution,

Raoult's law is obeyed for entire concentration range and temperatures
The enthalpy change of mixing is zero $\triangle H_{mix}=0.$
The volume of mixing is zero. \triangle $V_{mix}=0$
But the entropy of mixing is not equal to zero. $\triangle S_{mix}\neq0$

Option D is correct.

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