Question

$2.5 \, \text{g}$ of a non-volatile, non-electrolyte is dissolved in $100 \, \text{g}$ of water at $25^\circ \text{C}$. The solution showed a boiling point elevation by $2^\circ \text{C}$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is ______ mm of Hg (nearest integer).
(Given: Molal boiling point elevation constant of water ($K_b$) = $0.52 \, \text{K} \cdot \text{kg} \cdot \text{mol}^{-1}$,1 atm pressure = $760 \, \text{mm of Hg}$, molar mass of water = $18 \, \text{g mol}^{-1}$)

Answer 1

Determine the Molality of the Solution:

The boiling point elevation $\Delta T_b$ is related to molality ($m$) as follows: $\Delta T_b = K_b \times m$

Given:

$\Delta T_b = 2^\circ C, \quad K_b = 0.52 \, \text{K kg mol}^{-1}$ $m = \frac{\Delta T_b}{K_b} = \frac{2}{0.52} \approx 3.85 \, \text{mol kg}^{-1}$

Calculate the Moles of Solute:

Since molality $m$ is defined as moles of solute per kilogram of solvent: $\text{moles of solute} = m \times \text{mass of solvent (in kg)}$ Given that the mass of solvent (water) is 100 g or 0.1 kg: $\text{moles of solute} = 3.85 \times 0.1 = 0.385 \, \text{moles}$

Determine the Molar Mass of the Solute:

Given mass of solute = 2.5 g, $\text{Molar mass of solute} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{2.5}{0.385} \approx 6.49 \, \text{g/mol}$

Calculate the Vapour Pressure Lowering:

The vapour pressure lowering $\Delta P$ is given by: $\Delta P = P^0 \times \frac{\text{moles of solute}}{\text{moles of solvent}}$ where $P^0 = 760 \, \text{mm Hg}$ and moles of solvent (water) = $\frac{100}{18} \approx 5.56 \, \text{moles}.$

Calculate $\Delta P$:

$\Delta P = 760 \times \frac{0.385}{5.56} \approx 52.61 \, \text{mm Hg}$

Calculate the Vapour Pressure of the Solution:

$P_{\text{solution}} = P^0 - \Delta P = 760 - 52.61 \approx 707 \, \text{mm Hg}$

Conclusion:

The vapour pressure of the resulting aqueous solution is approximately $707 \, \text{mm Hg}$.