Solutions — Page 7 | NEET Chemistry

Q61

The freezing point of a diluted milk sample is found to be –0.2oC, while it should have been –0.5oC for pure milk. How much water has been added to pure milk to make the diluted sample?

A 1 cup of water to 2 cups of pure milk

B 2 cups of water to 3 cups of pure milk

C 3 cups of water to 2 cups of pure milk

D 1 cup of water to 3 cups of pure milk

Correct Answer

Option C

Solution

We know,

\Delta

Tf = i $\times$ kf $\times$ m $\therefore$

\Delta

TfDil. Milk = (1) $\times$ kf $\times$ mdil = 0.2 ...(1)

\Delta

TfPure Milk = (1) $\times$ kf $\times$ mpure = 0.5 ...(2) From (1) and (2), we get

{{{m_{pure}}} \over {{m_{dil}}}} = {{0.5} \over {0.2}}

=

{5 \over 2}

....(

3) Assume moles of fat in both the milk samples = n Weight of milk = w1 and weight of water = w0 Molality of pure milk, mpure =

{n \over {{w_1}}} \times 1000

....(4) Molality of diluted milk, mdil. =

{n \over {{w_1} + {w_0}}} \times 1000

....(5) From (4) and (5), we get

{{{m_{pure}}} \over {{m_{dil}}}} = {{{w_1} + {w_0}} \over {{w_1}}}

Using equation (3), we get

{5 \over 2} = {{{w_1} + {w_0}} \over {{w_1}}}

$\Rightarrow$ 5w1 = 2w1 + 2w0 $\Rightarrow$ 3w1 = 2w0 $\Rightarrow$

{{{w_1}} \over {{w_0}}} = {2 \over 3}

Which means 3 cups of water has been added to 2 cups of pure milk.

Q62

Match List I with List II .tg .tg List I List II A. van't Hoff factor, i I. Cryoscopic constant B.

\mathrm{k_f}

II. Isotonic solutions C. Solutions with same osmotic pressure III.

\mathrm{\dfrac{Normal\,molar\,mass}{Abnormal\,molar\,mass}}

D. Azeotropes IV. Solutions with same composition of vapour above it Choose the correct answer from the options given below :

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

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

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

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

Correct Answer

Option B

Solution

.tg .tg A. van't Hoff factor III.

\mathrm{{{Normal\,molar\,mass} \over {Abnormal\,molar\,mass}}}

B.

\mathrm{k_f}

I.

Cryoscopic constant C.

Solutions with same osmotic pressure II.

Isotonic solutions D.

Azeotropes IV.

Solutions with same composition of vapour above it

Q63

Given below are two statements: Statement (I) : Molal depression constant

\mathrm{K}_f

is given by

\dfrac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\mathrm{fus}}}

, where symbols have their usual meaning. Statement (II) :

\mathrm{K}_f

for benzene is less than the

\mathrm{K}_f

for water. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is correct but Statement II is incorrect

B Both Statement I and Statement II are incorrect

C Statement I is incorrect but Statement II is correct

D Both Statement I and Statement II are correct

Correct Answer

Option A

Solution

Statement-I Molar depression constant $\mathrm{k}_f=\dfrac{\mathrm{M}_1 \mathrm{RT}_f^2}{\Delta \mathrm{H}_{\text{fus }}}$

\begin{aligned} & \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_{\mathrm{f}}}{\left[\frac{\Delta \mathrm{H}_{\mathrm{fus}}}{\mathrm{~T}_{\mathrm{f}}}\right]} \\ & \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\text{fus }}} \end{aligned}

Hence statement-I is correct but $\mathrm{k}_{\mathrm{f}}$ for benzene $=5.12 \dfrac{{ }^{\circ} \mathrm{C}}{\text{ molal }}$ $\mathrm{k}_{\mathrm{f}}$ for water $=1.86 \dfrac{{ }^{\circ} \mathrm{C}}{\text{ molal }}$ Hence statement- II is incorrect

Q64

\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q)

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid isGiven:

K_f

(H2O) = 1.8 K kg mol−1, molality ≡ molarity

A

1.90 \times 10^{-3}

B

1.38 \times 10^{-3}

C

1.1 \times 10^{-2}

D

1.89 \times 10^{-1}

Correct Answer

Option B

Solution

Freezing Point Depression: $\Delta T_f = i \times K_f \times m$ Given: $\Delta T_f = 0.2 \, \text{°C}, \quad K_f = 1.8 \, \text{K kg mol}^{-1}, \quad m = 0.1 \, \text{m}$ Substituting the given values: $0.2 = i \times 1.8 \times 0.1$ Solving for $i$ : $i = \dfrac{0.2}{1.8 \times 0.1} = \dfrac{20}{18} = \dfrac{10}{9}$ Degree of Dissociation ( $\alpha$ ): For the reaction $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$ : $i = 1 + \alpha$ Given $i = \dfrac{10}{9}$ : $\dfrac{10}{9} = 1 + \alpha$ $\alpha = \dfrac{1}{9}$ Dissociation Constant ( $K_{eq}$ ): $\mathrm{K}_{eq} = \dfrac{[H^+][A^-]}{[HA]}$ At equilibrium: $[H^+] = [A^-] = \alpha \times C = \dfrac{1}{9} \times 0.1$ $[HA] = 0.1 \times (1 - \alpha) = 0.1 \times \left(1 - \dfrac{1}{9}\right)$ Substituting these into $K_{eq}$ : $\mathrm{K}_{eq} = \dfrac{(0.1 \times \dfrac{1}{9})^2}{0.1 \times \left(1 - \dfrac{1}{9}\right)}$ Simplifying: $\mathrm{K}_{eq} = \dfrac{0.1 \times \left(\dfrac{1}{81}\right)}{0.1 \times \dfrac{8}{9}} = \dfrac{1}{720}$ Therefore: $\mathrm{K}_{eq} = 1.38 \times 10^{-3}$

Q65

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

Q66

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.

Q67

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)

Q68

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.

Q69

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.

Q70

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