Solutions

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

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

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

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

Depression in freezing point is given by $\Delta$T_f = K_fm

where $m = {{1000 \times {w_{solute}}} \over {{M_{solute}} \times {w_{solvent}}}}$

Substituting the values, we get

$\Delta {T_f} = 2 \times {{1000 \times 34.5} \over {46 \times 500}} = 3K$

Therefore, T_f = 270 K and $T_f^o$ = 273 K. Hence, option (c) indicates correct representation of the graph.

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

$\matrix{ {M{X_2}} & { \to {M^{2 + }} + } & {2X + } \cr {(1 - \alpha )} & \alpha & {2\alpha } \cr } $

Thus, after dissociation total number of moles formed (n) = 3.

Now, we know degree of dissociation is

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

I. (CH₃OH) : Surface tension decreases as concentration increases.

II. (KCl) : Surface tension increases with concentration for ionic salt.

III. [CH₃(CH₂)₁₁ OSO$_3^ - $ Na⁺] : It is an anionic detergent.

There is decrease in surface tension before micelle formation, and after CMC (Critical Micelle Concentration) is attained, no change in surface tension.

When intermolecular attraction between two components A and B in the mixture is same as between A and A or B and B, hence it is a case of ideal solution. When intermolecular attraction between A and B in a mixture is smaller than that between A and A or B and B, then mixture is more vaporised, bp is lowered. It is a case of positive deviation from Raoult’s law. When intermolecular attraction between A and B is higher than that between A and A or B and B, then mixture is less vaporised, bp is increased. It is a case of negative deviation.

(a) Methanol molecules (CH)₃OH are hydrogen bonded. In a mixture of CCl₄ and CH₃OH, extent of H-bonding is decreased. Mixture is more vaporised thus, positive deviation from Raoult’s law.

(b) Acetone molecules have higher intermolecular attraction due to dipole-dipole interaction. With CS₂ , this interaction is decreased thus, positive deviation.

(c) Mixture of benzene and toluene forms ideal solution.

(d) Phenol and aniline have higher interaction due to intermolecular H-bonding. Hence, negative deviation.

Using the boiling point elevation formula:

$ \Delta T_b = K_b \cdot m $,

where:

$ \Delta T_b $ is the elevation in boiling point,

$ K_b $ is the ebullioscopic constant,

$ m $ is the molality.

Given:

$ K_b = 0.76 \, \text{K} \, \text{kg} \, \text{mol}^{-1} $,

$ \Delta T_b = 2^\circ\text{C} $,

mass of solute $ = 2.5 \, \text{g} $,

mass of solvent (water) $ = 100 \, \text{g} = 0.1 \, \text{kg} $.

First, find the molality $ m $:

$ m = \frac{\Delta T_b}{K_b} = \frac{2}{0.76} = \frac{200}{76} \approx 2.63 \, \text{mol} \, \text{kg}^{-1} $.

Now calculate the moles of solute:

Given the molality $ m $,

$ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} $.

Thus,

$ \text{moles of solute} = m \times \text{mass of solvent in kg} = 2.63 \times 0.1 = 0.263 \, \text{mol} $.

Next, determine the molecular weight $ M $ of the solute:

$ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{2.5 \, \text{g}}{0.263 \, \text{mol}} \approx 9.51 \, \text{g/mol} $.

Using Raoult's Law for the vapour pressure of the solution:

$ P_{\text{solution}} = P_0 \cdot (1 - \chi_{\text{solute}}) $,

where:

$ P_0 $ is the vapour pressure of pure solvent,

$ \chi_{\text{solute}} $ is the mole fraction of the solute.

Approximate the mole fraction of the solute:

$ \chi_{\text{solute}} = \frac{\text{moles of solute}}{\nu_{\text{solvent}} + \nu_{\text{solute}}} \approx \frac{0.263}{5.55 + 0.263} \approx \frac{0.263}{5.813} \approx 0.0452 $,

where $\nu_{\text{solvent}}$ are the moles of water (approximately 5.55 mol for 100 g of water).

Therefore, the vapour pressure of the solution:

$ P_{\text{solution}} = 760 \times (1 - 0.0452) \approx 760 \times 0.9548 \approx 724 \, \text{mm Hg} $.

Thus, the vapour pressure of the solution is 724 mm of Hg, corresponding to Option A.

To find the freezing point of the solution, we will use the formula for freezing point depression, which is given by:

$ \Delta T_f = i \cdot K_f \cdot m $

where:

• $ \Delta T_f $ is the freezing point depression.


• $ i $ is the van't Hoff factor (number of particles the solute splits into or forms in solution).


• $ K_f $ is the cryoscopic constant (freezing point depression constant), for water it's 1.86 K kg mol^-1.


• $ m $ is the molality of the solution.

Given:

• Solute is $ K_3[Fe(CN)_6] $, and molecular weight (M) = 329 g/mol.


• Mass of solute = 0.1 g.


• Mass of solvent (water) = 100 g.

First, we calculate the molality $ m $ of the solution:

$ m = \frac{{\text{Moles of solute}}}{{\text{Kilograms of solvent}}} $

Moles of solute:

$ \text{Moles of solute} = \frac{{\text{Mass of solute}}}{{\text{Molecular weight}}} = \frac{0.1}{329} \text{mol} $

$ \text{Moles of solute} = 3.04 \times 10^{-4} \text{ mol} $

Kilograms of solvent:

$ \text{Kilograms of solvent} = 100 \text{ g} = 0.1 \text{ kg} $

Now molality:

$ m = \frac{3.04 \times 10^{-4} \text{ mol}}{0.1 \text{ kg}} = 0.00304 \text{ mol/kg} $

Next, calculate the van't Hoff factor $ i $:

• The solute $ K_3[Fe(CN)_6] $ disassociates into 4 ions: 3 $ K^+ $ and 1 $ [Fe(CN)_6]^{3-} $. Therefore, $ i = 4 $.

Substituting these into the freezing point depression formula:

$ \Delta T_f = i \cdot K_f \cdot m = 4 \cdot 1.86 \cdot 0.00304 $

$ \Delta T_f = 0.02259 \text{ K} $

Convert to degrees Celsius and remember that this will lower the freezing point, so it should be negative (since the freezing point depression means the solution freezes at a lower temperature than pure solvent):

$ \Delta T_f = -0.02259 \text{ °C} $

So, $ \Delta T_f = -2.259 \times 10^{-2} \text{ °C} $

From the given options, the closest to this calculated value of $ \Delta T_f $ is:

Option A: -2.3 $\times$ 10^-2oC. Hence, Option A is the correct answer.

According to Henry's law, we have

${p_{{N_2}}} = {K_H}{x_{{N_2}}}$

$0.8 \times 5 = 1 \times {10^5} \times {x_{{N_2}}}$

${x_{{N_2}}} = 4 \times {10^{ - 5}}$

Thus one mole of solution contains $4 \times {10^{ - 5}}$ moles of N$_2$ and $1 - 4 \times {10^{ - 5}} \approx 1$ mole of water. Therefore, N$_2$ from air dissolved in 10 moles of water is $10 \times 4 \times {10^{ - 5}}$ moles $ = 4 \times {10^{ - 4}}$.