2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25°C. The solution showed a boiling point elevation by 2°C. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is ______ mm of Hg (nearest integer). [Given: Kb = 0.52 K·kgmol⁻¹, 1 atm = 760 mm Hg, molar mass of water = 18 g mol⁻¹]

The solute is non-volatile and non-electrolytic, hence the boiling-point elevation formula is

$$\Delta T_b = K_b\,m \quad -(1)$$
where $$m$$ is the molality of the solution.

Given $$\Delta T_b = 2^{\circ}\text{C}$$ and $$K_b = 0.52\ \text{K·kg mol}^{-1}$$, from $$(1)$$

$$m = \frac{\Delta T_b}{K_b} = \frac{2}{0.52} \approx 3.85\ \text{mol kg}^{-1} \quad -(2)$$

The mass of water (solvent) is 100 g = 0.100 kg.
Hence the number of moles of solute is

$$n_{\text{solute}} = m \times \text{mass of solvent (kg)}$$ $$= 3.85 \times 0.100 \approx 0.385\ \text{mol} \quad -(3)$$

Moles of water

$$n_{\text{water}} = \frac{100\ \text{g}}{18\ \text{g mol}^{-1}} \approx 5.56\ \text{mol} \quad -(4)$$

Because the solute concentration is very small compared with that of the solvent, the mole fraction of the solute can be approximated by

$$x_{\text{solute}} \approx \frac{n_{\text{solute}}}{n_{\text{water}}}$$

Substituting $$(3)$$ and $$(4)$$:

$$x_{\text{solute}} \approx \frac{0.385}{5.56} \approx 0.069 \quad -(5)$$

For a non-volatile solute, the relative lowering of vapour pressure is

$$\frac{P^\circ - P}{P^\circ} = x_{\text{solute}} \quad -(6)$$

where $$P^\circ$$ is the vapour pressure of pure solvent and $$P$$ is that of the solution.

At the normal boiling point of water (100 °C), $$P^\circ = 1\ \text{atm} = 760\ \text{mm Hg}$$. Using $$(5)$$ in $$(6)$$:

$$\frac{760 - P}{760} = 0.069$$ $$P = 760\,(1 - 0.069) \approx 760 \times 0.931 \approx 707\ \text{mm Hg}$$

Nearest integer:

Vapour pressure of the solution = $$707\ \text{mm Hg}$$.