Solutions — Page 8 | JEE Chemistry

Q71

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)

Q72

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.

Q73

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 $5.12 \, \text{K kg/mol}$ .

Q74

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 $80 \%$ .

Q75

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.

Q76

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.

Q77

Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is: (xM = Mole fraction of 'M' in solution ; xN = Mole fraction of 'N' in solution ; yM = Mole fraction of 'M' in vapour phase ; yN = Mole fraction of 'N' in vapour phase)

A

{{{x_M}} \over {{x_N}}} < {{{y_N}} \over {{y_N}}}

B (xM – yM) < (xN – yN)

C

{{{x_M}} \over {{x_N}}} = {{{y_N}} \over {{y_N}}}

D

{{{x_M}} \over {{x_N}}} > {{{y_M}} \over {{y_N}}}

Correct Answer

Option D

Solution

P_M^0

= 450 mmHg and

P_N^0

= 700 mmHg $\therefore$

P_M^0

<

P_N^0

Also we know,

P_M

=

P_M^0

X_M

=

Y_M

P_T

$\Rightarrow$

P_M^0

=

{{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)

P_N

=

P_N^0

X_N

=

Y_N

P_T

$\Rightarrow$

P_N^0

=

{{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)

As,

P_M^0

<

P_N^0

$\Rightarrow$

{{{Y_M}} \over {{X_M}}}\left( {{P_T}} \right)

<

{{{Y_N}} \over {{X_N}}}\left( {{P_T}} \right)

$\Rightarrow$

{{{Y_M}} \over {{X_M}}}

<

{{{Y_N}} \over {{X_N}}}

$\Rightarrow$

{{{Y_M}} \over {{Y_N}}}

<

{{{X_M}} \over {{X_N}}}

Q78

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be

A 360

B 350

C 300

D 700

Correct Answer

Option B

Solution

P_A^ \circ = ?,\,\,

Given

P_B^ \circ = 200mm,\,\,{x_A} = 0.6

{x_B} = 1 - 0.6 = 0.4,\,\,P = 290

P = {P_A} + {P_B} = P_A^ \circ {x_A} + P_B^ \circ {x_B}

\Rightarrow 290 = P_A^ \circ \times 0.6 + 200 \times 0.4

$\therefore$

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

P_A^ \circ = 350mm

Q79

An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :

A 0.33

B 0.50

C 0.67

D 0.80

Correct Answer

Option B

Solution

Let us assume that degree of dissociation is $\alpha$ .

\begin{array}{lll}{M{X_2}} & { \to {M^{2 + }} + } & {2X + } \\ {(1 - \alpha )} & \alpha & {2\alpha } \end{array}

Thus, after dissociation total number of moles formed (n) = 3. Now, we know degree of dissociation is

\alpha = {{i - 1} \over {n - 1}} = {{2 - 1} \over {3 - 1}} = 0.50

Q80

The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:

A 64

B 128

C 488

D 32

Correct Answer

Option A

Solution

Using relation,

{{{P^ \circ } - {P_s}} \over {{P_s}}} = {{{w_2}{M_1}} \over {{w_1}{M_2}}}

where

{w_1},

{M_1} =

mass in

g

and mol. mass of solvent

{w_2},{M_2} =

mass in

g

and mol. mass of solute Let

{M_2} = x

{P^ \circ } = 185\,\,torr;\,\,{P_s} = 183\,torr

{{185 - 183} \over {183}} = {{1.2 \times 58} \over {100x}}

(Mol. mass of acetone

=58

)

x=64

$\therefore$

\,\,\,\,\,

Molar mass of substance

=64