Solutions — Page 9 | NEET Chemistry

Q81

At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be. (molar mass of urea = 60 g mol–1)

A 0.031 mmHg

B 0.017 mmHg

C 0.028 mmHg

D 0.027 mmHg

Correct Answer

Option B

Solution

Given that, wsolute = wurea = 0.6 gm wsolvent = wH2O = 360 gm po = 35 We know, lowering of vapour pressure

\Delta

p = xsolute $\times$ po =

{{{n_{urea}}} \over {{n_{urea}} + {n_{{H_2}O}}}}

$\times$ po =

{{{{0.6} \over {60}}} \over {{{0.6} \over {60}} + {{360} \over {18}}}}

$\times$ 35 =

{{{{10}^{ - 2}}} \over {20}} \times 35

= 0.017

Q82

The depression in freezing point observed for a formic acid solution of concentration

0.5 \mathrm{~mL} \mathrm{~L}^{-1}

is

0.0405^{\circ} \mathrm{C}

. Density of formic acid is

1.05 \mathrm{~g} \mathrm{~mL}^{-1}

. The Van't Hoff factor of the formic acid solution is nearly : (Given for water

\mathrm{k}_{\mathrm{f}}=1.86\, \mathrm{k} \,\mathrm{kg}\,\mathrm{mol}^{-1}

)

A 0.8

B 1.1

C 1.9

D 2.4

Correct Answer

Option C

Solution

\Delta \mathrm{T}_{\mathrm{f}}

of formic acid

=0.0405^{\circ} \mathrm{C}

Concentration

=0.5 \mathrm{~mL} / \mathrm{L}

and density

=1.05 \mathrm{~g} / \mathrm{mL}

$\therefore$ Mass of formic acid in solution

=1.05 \times 0.5 \mathrm{~g}

=0.525 \mathrm{~g}

$\therefore$ According to Van't Hoff equation,

\begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m} \\ &0.0405=\mathrm{i} \times 1.86 \times \frac{0.525}{46 \times 1} \end{aligned}

(Assuming mass of

1 \mathrm{~L}

water

=\mathrm{kg}

)

\mathrm{i}=\frac{0.0405 \times 46}{1.86 \times 0.525}=1.89 \approx 1.9

Q83

If

\alpha

is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for calculating the molecular mass is :

A 1 +

\alpha

B 1 +

2\alpha

C 1 -

\alpha

D 1 -

2\alpha

Correct Answer

Option B

Solution

N{a_2}\,S{O_4}\,\rightleftharpoons\mathop {2N{a^ + }}\limits_{2\alpha } \,\, + \,\,\mathop {SO_4^{ - - }}\limits_\alpha

Vant. Hoff's factor

\,\,\,i = {{1 - \alpha + 2\alpha + \alpha } \over 1} = 1 + 2\alpha

Q84

A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)

A 0.162

B 0.675

C 0.325

D 0.486

Correct Answer

Option C

Solution

Mole fractions can be calculated as

{n_{C{H_2}C{l_2}}} = {{8.5} \over {8.5}} = 0.1

and

{n_{CHC{l_3}}} = {{11.9} \over {119.5}} = 0.1

We know that

{p_{total}} = p_{C{H_2}C{l_2}}^o \times \,{x_{C{H_2}C{l_2}}} + p_{CHC{l_3}}^o \times \,{x_{CHC{l_3}}}

= 415 \times 0.10 + 200 \times 0.1 = 61.5

Mole fraction of CHCl3 in vapour form can be calculated as

{p_{total}} = p_{CHC{l_3}}^o \times \,{x_{CHC{l_3}}}

61.5 = 200 \times \,{x_{CHC{l_3}}}

{x_{CHC{l_3}}} = {{61.5} \over {200}} = 0.3075

Q85

At 80oC, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80oC and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

A 52 mol percent

B 34 mol percent

C 48 mol percent

D 50 mol percent

Correct Answer

Option D

Solution

At

1

atmospheric pressure the boiling point of mixture is

{80^ \circ }C.

At boiling point the vapour pressure of mixture,

{P_T} = 1

atmosphere

= 760\,mm\,Hg.

Using the relation,

{P_T} = P_A^ \circ {X_A} + P_B^ \circ {X_B},\,\,

we get

{P_T} = 520{X_A} + 1000\left( {1 - {X_A}} \right)

\left\{ \, \right.

P_A^ \circ = 520mm\,\,Hg,

\,\,\,\,\,\,\,\,\,\,

\left. {P_B^ \circ = 1000\,mm\,Hg,\,{X_A} + {X_B} = 1} \right\}

or

\,\,\,\,\,\,\,\,\,\,

760 = 520{X_A} + 1000 - 1000{X_A}\,\,

or

\,\,\,\,\,\,\,\,\,\,

480{X_A} = 240

or

\,\,\,\,\,\,\,\,\,\,

{X_A} = {{240} \over {480}} = {1 \over 2}

or

\,\,\,\,\,\,\,\,\,\,

50

mol. percent i.e., The correct answer is

(d)

Q86

The vapour pressure of water at 20oC is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20oC, the vapour pressure of the resulting solution will be

A 17.675 mm Hg

B 15.750 mm Hg

C 16.500 mm Hg

D 17.325 mm Hg

Correct Answer

Option D

Solution

NOTE : On addition of glucose to water, vapour pressure of water will decrease.

The vapour pressure of a solution of glucose in water can be calculated using the relation

{{{P^ \circ } - {P_S}} \over {{P_S}}} = {{Moles\,\,of\,\,glu\cos e\,\,in\,\,solution} \over {moles\,\,of\,\,water\,\,in\,\,solution}}

or

\,\,\,\,\,\,\,\,\,

{{17.5 - {P_S}} \over {{P_S}}} = {{18/180} \over {178.2/18}}

[as

\,\,\,\,\,\,\,\,\,

{P^ \circ } = 17.5

] or

\,\,\,\,\,\,\,\,\,

17.5 - {P_S} = {{0.1 \times {P_S}} \over {9.9}}

or

\,\,\,\,\,\,\,\,\,

{P_S} = 17.325\,mm\,Hg.

Hence (d) is correct answer.

Q87

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass

256 \mathrm{~g} \mathrm{~mol}^{-1}

) and the decrease in freezing point is 0.40 K ?

A

4.43 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

B

3.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

C

5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

D

1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

Correct Answer

Option C

Solution

To find the freezing point depression constant ( $K_f$ ) of the solvent, we use the formula for freezing point depression: $\Delta T_f = K_f \cdot m$ Given: The decrease in freezing point $\Delta T_f$ is 0.40 K.

The mass of the solute is 1 g and its molar mass is 256 g/mol.

The mass of the solvent is 50 g (or 0.050 kg).

First, calculate the molality ( $m$ ): Molality is defined as the moles of solute per kilogram of solvent.

Calculate moles of solute: $\text{Moles of solute} = \dfrac{1 \, \text{g}}{256 \, \text{g/mol}} = \dfrac{1}{256} \, \text{mol}$ Calculate molality ( $m$ ): $m = \dfrac{\dfrac{1}{256} \, \text{mol}}{0.050 \, \text{kg}} = \dfrac{1}{256 \times 0.050} \, \text{mol/kg}$ Now, substitute into the formula to find $K_f$ : $0.4 = K_f \cdot \dfrac{1}{256 \times 0.050}$ Solving for $K_f$ : $K_f = 0.4 \cdot 256 \times 0.050 = 5.12 \, \text{K kg/mol}$ Thus, the freezing point depression constant of the solvent is $5.12 \, \text{K kg/mol}$ .

Q88

Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2

^\circ

C. The percentage association of solute A in water, is : [Given : Molar mass of A = 93 g mol

-

1. Molal depression constant of water is 1.86 K kg mol

-

1.]

A 50%

B 60%

C 70%

D 80%

Correct Answer

Option D

Solution

Since, $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f} m}$

\begin{aligned} &m=\frac{0.7}{93} \times \frac{1000}{42} \\\\ &0.2=i \times 1.86 \times \frac{0.7 \times 1000}{93 \times 42} \\\\ &i=0.6 \\\\ &\alpha=\frac{i-1}{\frac{1}{n}-1}=\frac{0.6-1}{\frac{1}{2}-1}=0.8 \end{aligned}

Hence, the percentage association of solute $A$ is $80 \%$ .

Q89

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : At 10

^\circ

C, the density of a 5 M solution of KCl [atomic masses of K & Cl are 39 & 35.5 g mol

-

1 respectively], is 'x' g ml

-

1. The solution is cooled to

-

21

^\circ

C. The molality of the solution will remain unchanged. Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature. In the light of the above statements, choose the correct answer from the options given below :

A Both (A) and (R) are true and (R) is the correct explanation of (A).

B Both (A) and (R) are true but (R) is not the correct explanation of (A).

C (A) is true but (R) is false.

D (A) is false but (R) is true.

Correct Answer

Option A

Solution

Molality and Mass are temperature Independent so on changing temp., molality and mass remain unchanged.

Q90

A solution of two miscible liquids showing negative deviation from Raoult's law will have :

A increased vapour pressure, increased boiling point

B increased vapour pressure, decreased boiling point

C decreased vapour pressure, decreased boiling point

D decreased vapour pressure, increased boiling point

Correct Answer

Option D

Solution

Negative Deviation from Raoult's Law Negative deviation means the intermolecular forces of attraction between the molecules of the two liquids (A-B) are stronger than the forces between molecules of the pure liquids (A-A and B-B).

This stronger attraction makes it harder for molecules to escape into the vapor phase.

Effect on Vapor Pressure and Boiling Point Vapor Pressure : Since the molecules are held more tightly, the vapor pressure of the solution will be lower than expected from Raoult's law.

Decreased vapor pressure.

Boiling Point : A lower vapor pressure means you need to increase the temperature further to reach atmospheric pressure, where boiling occurs.

Therefore, the boiling point of the solution will be higher than expected.

Increased boiling point.

Answer The correct answer is Option D: decreased vapor pressure, increased boiling point.