Calculate the Vapour Pressure of Solution M

To find the vapour pressure of the solution $\mathbf{M}$, we analyze the composition of the solution using Raoult's Law and colligative properties.

1. Determination of the Composition of Solution M

Let the solution $\mathbf{M}$ be an aqueous solution containing a non-volatile, non-electrolyte solute.

From the context of the linked problem (Question 69), water is added to solution $\mathbf{M}$ until the mole fraction of water becomes $X_{\text{water}}' = 0.9$ (which means the mole fraction of the solute is $X_{\text{solute}}' = 1 - 0.9 = 0.1$).

The molality $m'$ of this diluted solution is:
$m' = \frac{n_{\text{solute}}}{\text{mass of water (in kg)}} = \frac{X_{\text{solute}}'}{X_{\text{water}}' \times M_{\text{water}}} = \frac{0.1}{0.9 \times 0.018\text{ kg}\cdot\text{mol}^{-1}} \approx 6.173\text{ mol}\cdot\text{kg}^{-1}$

The elevation in boiling point for this diluted solution is:
$\Delta T_b = K_b \cdot m' = 0.52\text{ K}\cdot\text{kg}\cdot\text{mol}^{-1} \times 6.173\text{ mol}\cdot\text{kg}^{-1} \approx 3.21\text{ K}$

Thus, the boiling point of the diluted solution is:
$T_b = T_b^\circ + \Delta T_b = 373.0\text{ K} + 3.2\text{ K} = 376.2\text{ K}$
which matches the result given in Question 69.

In the original solution $\mathbf{M}$, the mole fraction of water is:
$X_{\text{water}} = 0.8 \quad \text{and} \quad X_{\text{solute}} = 0.2$

(Note: Prior to the addition of water, the molality of solution $\mathbf{M}$ is $m = \frac{0.2}{0.8 \times 0.018} \approx 13.89\text{ mol}\cdot\text{kg}^{-1}$, giving $\Delta T_b = 0.52 \times 13.89 \approx 7.22\text{ K}$, and a boiling point of $373.15 + 7.22 \approx 380.4\text{ K}$, which corresponds to option (A) of Question 69).

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2. Calculation of the Vapour Pressure of Solution M

According to Raoult's Law for a solution containing a non-volatile solute:
$P_{\mathbf{M}} = X_{\text{water}} \cdot P^\circ_{\text{water}}$

Given the vapour pressure of pure water at the specified temperature:
$P^\circ_{\text{water}} = 45.0\text{ mm Hg}$

Substituting the mole fraction of water in solution $\mathbf{M}$:
$P_{\mathbf{M}} = 0.8 \times 45.0\text{ mm Hg} = 36.0\text{ mm Hg}$

Therefore, the vapour pressure of the solution $\mathbf{M}$ is $36.0\text{ mm Hg}$.

Correct Option: (B)