Solutions — Page 3 | JEE Chemistry

Q21

Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is

A 1 : 4

B 1 : 0.25

C 1 : 0.20

D 1 : 5

Correct Answer

Option B

Solution

\Delta T_{f}=i k_{f} \times m

\frac{\Delta T_{\mathrm{f}(\mathrm{A})}}{\Delta \mathrm{T}_{\mathrm{f}(\mathrm{B})}}=\frac{1}{4}

\frac{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{A}}} \times 1}{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{B}}} \times 1}=\frac{1}{4}

\frac{M_{B}}{M_{A}}=\frac{1}{4}

M_{A}: M_{B}=4: 1

Q22

In the depression of freezing point experiment A. Vapour pressure of the solution is less than that of pure solvent B. Vapour pressure of the solution is more than that of pure solvent C. Only solute molecules solidify at the freezing point D. Only solvent molecules solidify at the freezing point Choose the most appropriate answer from the options given below :

A A and C only

B A only

C A and D only

D B and C only

Correct Answer

Option C

Solution

In the depression of freezing point experiment only solvent molecules solidify and vapour pressure of solution decreases as some of the surface area is occupied by solute molecules, so less number of molecules will go in vapour form.

Q23

Identify the mixture that shows positive deviations from Raoult's Law

A

\left(\mathrm{CH}_3\right)_2 \mathrm{CO}+\mathrm{CS}_2

B

\left(\mathrm{CH}_3\right)_2 \mathrm{CO}+\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2

C

\mathrm{CHCl}_3+\mathrm{C}_6 \mathrm{H}_6

D

\mathrm{CHCl}_3+\left(\mathrm{CH}_3\right)_2 \mathrm{CO}

Correct Answer

Option A

Solution

\left(\mathrm{CH}_3\right)_2 \mathrm{CO}+\mathrm{CS}_2

Exibits positive deviations from Raoult's Law

Q24

What happens to freezing point of benzene when small quantity of napthalene is added to benzene?

A Increases

B Decreases

C Remains unchanged

D First decreases and then increases

Correct Answer

Option B

Solution

When a small quantity of naphthalene is added to benzene, the freezing point of benzene decreases.

This phenomenon is due to the colligative property known as freezing point depression.

The addition of a solute, such as naphthalene, disrupts the orderly arrangement of solvent molecules in the solid phase, thereby lowering the freezing point.

Mathematically, the decrease in freezing point ( $\Delta T_f$ ) can be represented by the equation:

\Delta T_f = i \cdot K_f \cdot m

where: $i$ is the van't Hoff factor (which is 1 for naphthalene as it does not dissociate in solution), $K_f$ is the cryoscopic constant of the solvent (benzene), $m$ is the molality of the solution.

Thus, the correct option is: Option B: Decreases

Q25

1.24 g of AX2 (molar mass 124 g mol−1) is dissolved in 1 kg of water to form a solution with boiling point of 100.015°C, while 25.4 g of AY2 (molar mass 250 g mol−1) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.Kb(H2O) = 0.52 kg mol−1Which of the following is correct?

A AX2 and AY2 (both) are completely unionised.

B AX2 and AY2 (both) are fully ionised.

C AX2 is completely unionised while AY2 is fully ionised.

D AX2 is fully ionised while AY2 is completely unionised.

Correct Answer

Option D

Solution

Mass of $A x_2=1.24 \mathrm{~g}$ (solute) Molarmass of $A X_2=124 \mathrm{~g} \mathrm{~mol}^{-1}$ Mass of water $=1 \mathrm{~kg}$ (solvent.)

Boiling point of water $=100^{\circ} \mathrm{C}$ Boiling point of water after adding solute $A X_2=100.0156^{\circ} \mathrm{C}$ Mass of $A Y_2=25.4 \mathrm{~g}$ (solute) Molarmass of $A Y_2=250 \mathrm{~g~mol}^{-1}$ Mass of water $=2 \mathrm{~kg}$ (Solvent) Boiling point of water $\mp 100^{\circ} \mathrm{C}$ Boiling point of water after adding solute $A y_2=100.0260^{\circ} \mathrm{C}$ $\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1}=0.520^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}$ The ionisation of $A x_2$ and $A Y_2$ can be determined by calculating Van't Hoff factor.

The phenomenon given in the question is elevation in boiling point.

The boiling point of solvent increases when another compound (solute) is added to it.

Relation: $\Delta T_b=K_b \cdot i \cdot m$

\Delta T_b=T_b-T_b^0

$\Delta T_b \rightarrow$ Boiling point elevation $T_b \rightarrow$ Boiling point of solution (solvent + solute) $T_b^0 \rightarrow$ Boiling point of solvent $K_b \rightarrow$ Molal elevation constant $i \rightarrow$ Van't Hoff factor

m \rightarrow \text{ Molality }=\frac{\text{ Number of moles }}{\mathrm{kg} \text{ of solvent }}, \text{ Moles }=\frac{\text{ Mass }}{\text{ Molarmass }}

For $A X_2$ and $A Y_2$ (solute), Van't Hoff factor represents how many particles a solute dissociates into when dissolved in a solvent (water).

For non-electrolytes, $i=1$

\begin{aligned} &\begin{aligned} & A X_2: \\ & \text{ Moles }=\frac{\text{ mass }}{\text{ molarmass }}=\frac{1.249}{124 \mathrm{~g~mol^{-1}}}=0.01 \mathrm{~mol} \\ & \text{ Molality }=\frac{\text{ Moles }}{k_g \text{ of solvent }}=\frac{0.01 \mathrm{~mol}}{1 \mathrm{~kg}}=0.01 \mathrm{~mol} \mathrm{~kg}^{-1} \\ & \Delta T_b=K_b \times i \times m \\ & i=\frac{\Delta T_b}{k_b \times m} \\ &=\frac{T_b-T_b^0}{K_b \times m} \end{aligned}\\ &\text{ Substitute values as., } \end{aligned}

\begin{aligned} i & =\frac{\left(100.0156^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}\right)}{0.52^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.01 \mathrm{~mol} \mathrm{~kg}^{-1}} \\ & =\frac{0.0156^\circ \mathrm{C}}{0.52 \times 0.01{ }^{\circ} \mathrm{C}}=3 \end{aligned}

$i=3$ means, there are 3 particles in solution after $A X_2$ dissolved in water.

A x_2 \rightarrow 1 A^{+}+2 X^{-} \quad(1+2=3)

\begin{aligned} &A x_2 \text{ is completely ionised. }\\ &A Y_2: \end{aligned}

\begin{aligned} &\begin{aligned} \text{ Moles } & =\frac{\text{ Mass }}{\text{ Molarmass }}=\frac{25.4 \mathrm{~g}}{250 \mathrm{g~mol}}=0.1016 \mathrm{~mol} \\ \text{ Molality } & =\frac{\text{ Moles }}{\mathrm{kgof}^{-1} \text{ solvent }}=\frac{0.1016 \mathrm{~mol}}{2 \mathrm{~kg}}=0.050 .8 \mathrm{~mol} \mathrm{~kg}^{-1} \\ \Delta T_b & =k_b \times \mathrm{l}^2 \times \mathrm{m} \\ i & =\frac{\Delta T_b}{k_b \times m} \\ & =\frac{T_b-T_b^0}{k_b \times m} \end{aligned}\\ &\text{ Substitute values as, } \end{aligned}

i = {{(100.0260^\circ C - 100^\circ C)} \over {0.52^\circ C\,kg\,mo{l^{ - 1}} \times 0.0508\,mol\,k{g^{ - 1}}}}

$=\dfrac{0.0260 ^\circ \mathrm{C}}{0.52 \times 0.0508 ^\circ \mathrm{C}}$ $= 0.98$ $\approx 1$ $i=1$ means $A Y_2$ is completely unionised. $A y_2$ not give ionised particles when dissolved in water.

So, $A X_2$ is completely ionised and $A Y_2$ is completely unionised.

Answer: Option 4) $A x_2$ is fully ionised, $A Y_2$ is completely unionised.

Q26

When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?

A 0.2

B 0.4

C 0.8

D 0.6

Correct Answer

Option D

Solution

When a non-volatile solute is added to a solvent, it causes the vapour pressure of the solvent to decrease.

In this scenario, when the vapour pressure decreases by 10 mm of Hg, the mole fraction of the solute in the solution is 0.2.

By understanding the relationship between vapour pressure change and mole fraction, we see that: The change in vapour pressure ( $P^{\circ} - P$ ) is directly proportional to the mole fraction of the solute ( $X_{\text{solute}}$ ).

Therefore, if a 10 mm of Hg decrease corresponds to a mole fraction of 0.2, then a 20 mm of Hg decrease would correspond to a mole fraction of 0.4.

To find the mole fraction of the solvent ( $X_{\text{solvent}}$ ), we use the formula: $X_{\text{solvent}} = 1 - X_{\text{solute}}$ Substituting the value we found: $X_{\text{solvent}} = 1 - 0.4 = 0.6$ Thus, when the vapour pressure decreases by 20 mm of Hg, the mole fraction of the solvent is 0.6.

Q27

The size of a raw mango shrinks to a much smaller size when kept in a concentrated salt solution. Which one of the following processes can explain this?

A Osmosis

B Reverse osmosis

C Diffusion

D Dialysis

Correct Answer

Option A

Solution

Raw mango shrink in salt solution due to net transfer of water molecules from mango to salt solution due to phenomenon of osmosis.

Q28

On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 and of octane = 114 g mol–1)

A 72.0 kPa

B 36.1 kPa

C 96.2 kPa

D 144.5 kPa

Correct Answer

Option A

Solution

{P_{Total}} = {P_A}^ \circ {x_A} + {P_B}^ \circ {x_B}

= {P^ \circ }_{Hep\tan e}\,{x_{hep\tan e}} + {P^ \circ }_{Oc\tan e}\,{x_{Oc\tan e}}

= 105 \times {{25/100} \over {{{25} \over {100}} + {{35} \over {114}}}} + 45 \times {{35/114} \over {{{25} \over {100}} + {{35} \over {114}}}}

= 105 \times {{0.25} \over {0.25 + 0.3}} + 45 \times {{0.3} \over {0.25 + 0.3}}

= {{105 \times 0.25} \over {0.55}} + {{45 \times 0.3} \over {0.55}}

= {{26.25 + 13.5} \over {0.55}} = 72\,kPa

Q29

Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is -

A 4A

B 2A

C 3A

D A

Correct Answer

Option C

Solution

For same freezing point, (

\Delta

Tf)X = (

\Delta

Tf)Y $\Rightarrow$ kf mx = kf my $\Rightarrow$

{{4 \times 1000} \over {A \times 96}} = {{12 \times 1000} \over {M \times 88}}

$\Rightarrow$ M = 3.27A

\simeq

3A

Q30

1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points,

{{\Delta {T_b}(A)} \over {\Delta {T_b}(B)}}

, is :

A 5 : 1

B 1 : 0.2

C 10 : 1

D 1 : 5

Correct Answer

Option D

Solution

Given,

{{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} = {1 \over 5}

\Delta {T_b}\left( A \right) = {k_b}\left( A \right) \times

Molality of A

\Delta {T_b}\left( B \right) = {k_b}\left( B \right) \times

Molality of B Here, Molality of A = Molality of B = m

{{\Delta {T_b}\left( A \right)} \over {\Delta {T_b}\left( B \right)}} = {{{{\left( {{k_b}} \right)}_A}} \over {{{\left( {{k_b}} \right)}_B}}} \times {m \over m}

=

{1 \over 5}