JEE Chemistry 2023 Question with Solution

Given: $80\%$ of $\mathrm{MgCl_2}$ is dissociated, so $\alpha = 0.8$. The solution is $1.0$ molal. Find: Vapour pressure of the solution.

MgCl_2 dissociates as

$$\mathrm{MgCl_2 \rightarrow Mg^{2+} + 2Cl^-}$$

So the number of ions formed is $n = 3$.

The van't Hoff factor is

$$i = 1 + \alpha(n-1)$$

Substituting,

$$i = 1 + 0.8(3-1) = 2.6$$

Therefore, effective molality is

$$= i \times 1.0 = 2.6$$

For a dilute solution,

$$x_{\text{solute}} \approx \frac{(2.6)(18)}{1000} = 0.0468$$

Hence,

$$x_{\text{solvent}} = 1 - 0.0468 = 0.9532$$

Using Raoult's law,

$$P_{\text{solution}} = x_{\text{solvent}} P^\circ_{\text{water}}$$

With $P^\circ_{\text{water}} = 50 \, \text{mm Hg}$,

$$P_{\text{solution}} = 0.9532 \times 50 = 47.66 \, \text{mm Hg}$$

Rounding to the nearest integer,

$$P_{\text{solution}} \approx 48 \, \text{mm Hg}$$

Therefore, the vapour pressure is $48 \, \text{mm Hg}$.

Given: Degree of dissociation of $\mathrm{MgCl_2}$ is $0.8$. Find: The nearest integer value of vapour pressure.

1. First calculate the van't Hoff factor:

$$i = 1 + \alpha(n-1)$$

Here $\alpha = 0.8$ and $n = 3$, so

$$i = 1 + 0.8 \times 2 = 2.6$$

2. Effective concentration of particles becomes $2.6$ times the given molality.

3. Approximate mole fraction of solute particles:

$$x_{\text{solute}} \approx \frac{2.6 \times 18}{1000} = 0.0468$$

Therefore,

$$x_{\text{solvent}} = 1 - 0.0468 = 0.9532$$

4. Apply Raoult's law:

$$P = x_{\text{solvent}} P^\circ$$

Using $P^\circ = 50 \, \text{mm Hg}$,

$$P = 0.9532 \times 50 = 47.66 \, \text{mm Hg}$$

Nearest integer value is $48$.

Therefore, the answer is $48$.