Solutions — Page 6 | JEE Chemistry

Q51

Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively :

A 200 and 300

B 300 and 400

C 400 and 600

D 500 and 600

Correct Answer

Option C

Solution

{P_{total}} = P_A^ \circ {X_A} + P_B^ \circ {X_B};

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

550 = P_A^ \circ \times {1 \over 4} + P_B^ \circ \times {3 \over 4}

P_A^ \circ + 3P_B^ \circ = 550 \times 4\,\,...\left( i \right)

In second case

{P_{total}} = P_A^ \circ \times {1 \over 5} + P_B^ \circ \times {4 \over 5}

P_A^ \circ + 4P_B^ \circ = 560 \times 5\,\,...\left( {ii} \right)

Subtract

(i)

from

(ii)

$\therefore$

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

P_B^ \circ = 560 \times 5 - 550 \times 4 = 600

As

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

P_A^ \circ = 400

Q52

Which of the following liquid pairs shows a positive deviation from Raoult’s law?

A Water – hydrochloric acid

B Acetone – chloroform

C Water – nitric acid

D Benzene – methanol

Correct Answer

Option D

Solution

NOTE : Positive deviations are shown by such solutions in which solvent-solvent and solute-solute interactions are stronger than the solvent interactions.

In such solution, the interactions among molecules becomes weaker.

Therefore their escaping tendency increases which results in the increase in their partial vapour pressures.

In a solutions of benzene and methanol there exists inter molecular

H-

bonding.

In this solution benzene molecules come between ethanol molecules which weaken intermolecular forces.

This results in increase in vapour pressure.

Q53

Match with .

List - I		List - II
(A)	Solution of chloroform and acetone	(I)	Minimum boiling azeotrope
(B)	Solution of ethanol and water	(II)	Dimerizes
(C)	Solution of benzene and toluene	(III)	Maximum boiling azeotrope
(D)	Solution of acetic acid in benzene	(IV)	ΔVmix = 0

A (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

B (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

C (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

D (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Correct Answer

Option A

Solution

To correctly match the items from List - I with those in List - II, let's analyze the characteristics of each solution: (A) Solution of chloroform and acetone: This solution exhibits negative deviation from Raoult's law.

It creates a maximum boiling azeotrope because the interactions between chloroform and acetone molecules are stronger than those in the pure components.

(B) Solution of ethanol and water: This solution shows positive deviation from Raoult's law, leading to the formation of a minimum boiling azeotrope.

The interactions between ethanol and water molecules are weaker than those in their pure states.

(C) Solution of benzene and toluene: This combination forms an ideal solution where Raoult’s law is obeyed across all concentrations.

Thus, the volume change upon mixing, denoted as $\Delta V_{\text{mix}}$ , is zero.

(D) Solution of acetic acid in benzene: In this mixture, acetic acid tends to dimerize.

The acetic acid molecules pair up, forming dimers, especially in non-polar solvents like benzene.

Based on these explanations, the correct matches are: (A) - (III) (B) - (I) (C) - (IV) (D) - (II)

Q54

Liquid A and B form an ideal solution. The vapour pressures of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If

x_A

and

x_B

are the mole fraction of A and B in solution while

y_A

and

y_B

are the mole fraction of A and B in vapour phase, then,

A $(x_A - y_A)

B

\dfrac{x_A}{x_B} = \dfrac{y_A}{y_B}

C $\frac{x_A}{x_B}

D

\dfrac{x_A}{x_B} > \dfrac{y_A}{y_B}

Correct Answer

Option D

Solution

Liquid A and B form an ideal solution.

The vapor pressures of pure liquids A and B are 350 mm Hg and 750 mm Hg, respectively, at the same temperature.

Here, $x_A$ and $x_B$ represent the mole fractions of A and B in the solution, and $y_A$ and $y_B$ are their mole fractions in the vapor phase.

Let’s begin by comparing the vapor pressures: $\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}$ \frac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}} The relationship between the mole fractions in the vapor phase and the solution can be expressed as: $\dfrac{y_A}{y_B} = \dfrac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}} \cdot \dfrac{x_A}{x_B}$ Since $\dfrac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}}$ \frac{\frac{y_A}{y_B}}{\frac{x_A}{x_B}} Which implies: $ \frac{y_A}{y_B} This indicates that the mole fraction ratio of A to B in the vapor phase is less than that in the solution.

Q55

Which one of the following aqueous solutions will exhibit highest boiling point?

A 0.01 M Na2SO4

B 0.015 M glucose

C 0.015 M urea

D 0.01 M KNO3

Correct Answer

Option A

Solution

As

\,\,\,\,\Delta {T_b}^b = {T_b} - {T_b}^o

Where

{T_b} = b.pt

of solution

T_b^o = b.pt

of solvent or

{T_b} = T_b^o + \Delta {T_b}

NOTE : Elevation in boiling point is a colligative property, which depends upon the no. of particles.

Thus greater the number of particles, greater is it elevation and hence greater will be its boiling point.

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

$ Since

N{a_2}S{O_4}\,\,

has maximum number of particles

(3)

hence has maximum boiling point.

Q56

The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :

A 37.5 g

B 75 g

C 150 g

D 50 g

Correct Answer

Option C

Solution

Molar mass of octane (C8 H18) = 8 $\times$ 12 + 18 = 114 g/mol Let, $\omega$ is the mass of solute.

Relative lowering vapour pressure,

{{\Delta P} \over P}

=

{{{\omega \over {50}}} \over {{\omega \over {50}} + {{114} \over {114}}}}

$\Rightarrow$

\,\,\,\,

{{75} \over {100}}

=

{{{\omega \over {50}}} \over {{\omega \over {50}} + 1}}

$\Rightarrow$

\,\,\,\,

{3 \over 4}

\left( {{\omega \over {50}} + 1} \right) = {\omega \over {50}}

$\Rightarrow$

\,\,\,\,

$\omega$ = 150 g

Q57

Which one of the following statements regarding Henry's law is not correct ?

A Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids

B Different gases have different KH (Henry's law constant) values at the same temperature.

C The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.

D The value of KH increases with increase of temperature and KH is function of the nature of the gas

Correct Answer

Option A

Solution

From Henry's law we know, Pgas = KH xg Where, Pgas = Pressure of undissolved gas xg = Mole fraction of gas dissolved into the liquid.

KH = Henry's Carnot.

When pressure is constant then, xg $\propto$

{1 \over {K{}_H}}

So. when KH is high then xg or solubility of gas is lower in the liquid. So, option (A) is wrong.

Q58

Among the following mixtures, dipole-dipole as the major interaction, is present in

A benzene and ethanol

B acetonitrile and acetone

C KCl and water

D benzene and carbon tetrachloride

Correct Answer

Option B

Solution

Acetonitrile

\mathop {\left( {C{H_3}} \right.}\limits^{\delta + } \,

\,\, - \,\,\,C \equiv \,\,

\mathop {\left. N \right)}\limits^{\delta - }

and acetone dipole-dipole interaction exist between them. Between

KCl

and water ion-dipole interaction is found and in Benzene - ethanol, Benzene is non polar but ethanol is polar so it create dipole - induced dipole interaction Benzene $-$ Carbon tetra chloride, both are non polar so one create instantaneous dipole and other gets induced by it.

Q59

Molecules of benzoic acid (C6H5COOH) dimerise in benzene. 'w' g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimmer in the solution is 80, then w is – (Its given that Kf = 5 K kg mol–1, Molar mass of benzoic acid = 122 g mol–1)

A 1.5 g

B 1.8 g

C 1.0 g

D 2.4 g

Correct Answer

Option D

Solution

We know,

\Delta

Tf = i Kf $\times$

{w \over M} \times {{1000} \over {{w_s}}}

i = 1 + $\alpha$

\left( {{1 \over n} - 1} \right)

Here Benzoic acid dimerise, so value of n = 2 $\therefore$ i = 1 + 0.8

\left( {{1 \over 2} - 1} \right)

= 1 $-$ 0.4 = 0.6 $\therefore$ 2 = 0.6 $\times$ 5 $\times$

{w \over {122}} \times {{1000} \over {30}}

$\Rightarrow$ w = 2.44 g

Q60

Boiling point of a

2 \%

aqueous solution of a non-volatile solute A is equal to the boiling point of

8 \%

aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is

A

\mathrm{M}_{\mathrm{A}}=4 \mathrm{M}_{\mathrm{B}}

B

\mathrm{M}_{\mathrm{B}}=4 \mathrm{M}_{\mathrm{A}}

C

\mathrm{M}_{\mathrm{A}}=8 \mathrm{M}_{\mathrm{B}}

D

\mathrm{M}_{\mathrm{B}}=8 \mathrm{M}_{\mathrm{A}}

Correct Answer

Option B

Solution

For $\mathbf{A}: 100 \,\mathrm{gm}$ solution $\rightarrow 2 \,\mathrm{gm}$ solute $\mathrm{A}$ $\therefore$ Molality $=\dfrac{2 / \mathrm{M}_{\mathrm{A}}}{0.098}$ For B : $100 \,\mathrm{gm}$ solution $\rightarrow 8 \,\mathrm{gm}$ solute $\mathrm{B}$

\begin{aligned} &\therefore \text{ Molality }=\frac{8 / \mathrm{M}_{\mathrm{B}}}{0.092} \\\\ &\because\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{A}}=\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{B}} \end{aligned}

$\therefore$ Molality of $\mathrm{A}=$ Molality of $\mathrm{B}$

\therefore \frac{2}{0.098 \mathrm{M}_{\mathrm{A}}}=\frac{8}{0.092 \mathrm{M}_{\mathrm{B}}}

\frac{2}{98} \times \frac{92}{8}=\frac{M_A}{M_B}

\frac{1}{4.261}=\frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}}

\therefore \mathrm{M}_{\mathrm{B}}=4.261 \times \mathrm{M}_{\mathrm{A}}