Vapour Pressure Calculation from Elevation in Boiling Point

To determine the vapour pressure of the given solution, we use the concepts of elevation in boiling point and Raoult's law for relative lowering of vapour pressure.

Step 1: Calculate the molality ($m$) of the solution

The elevation in boiling point ($\Delta T_b$) is given by: $\Delta T_b = K_b \times m$

Given:

• $\Delta T_b = 2^\circ\text{C} = 2\text{ K}$
• $K_b = 0.76\text{ K kg mol}^{-1}$

Substituting the values: $2 = 0.76 \times m$ $m = \frac{2}{0.76} = \frac{50}{19}\text{ mol kg}^{-1}$

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Step 2: Relate molality to the mole fraction of solute

According to Raoult's law, the relative lowering of vapour pressure for a dilute solution containing a non-volatile solute is given by: $\frac{P^\circ - P}{P^\circ} = X_{\text{solute}}$

Where:

• $P^\circ$ is the vapour pressure of pure water at its boiling point ($1\text{ atm} = 760\text{ mm Hg}$).
• $P$ is the vapour pressure of the solution.
• $X_{\text{solute}}$ is the mole fraction of the solute.

Since the solution is dilute ($n_{\text{solute}} \ll n_{\text{solvent}}$): $X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solvent}} + n_{\text{solute}}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}$

We can express this ratio in terms of molality ($m$) and the molar mass of solvent ($M_A = 18\text{ g mol}^{-1}$ for water): $X_{\text{solute}} \approx \frac{m \times M_A}{1000}$

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Step 3: Calculate the vapour pressure of the solution ($P$)

Substituting $X_{\text{solute}}$ into Raoult's law equation: $\frac{P^\circ - P}{P^\circ} = \frac{m \times M_A}{1000}$

$\frac{760 - P}{760} = \left(\frac{50}{19}\right) \times \frac{18}{1000}$

$\frac{760 - P}{760} = \frac{900}{19000} = \frac{9}{190}$

Solving for $(760 - P)$: $760 - P = 760 \times \frac{9}{190} = 4 \times 9 = 36\text{ mm Hg}$

$P = 760 - 36 = 724\text{ mm Hg}$

Therefore, the vapour pressure of the solution is $724\text{ mm Hg}$, which corresponds to Option A.