Question

Vapour pressure of pure water at 298 K is 24.8 mm Hg. Calculate the lowering in vapour pressure of an aqueous solution which freezes at –0.3°C. (K\(_f\) of water = 1.86 K kg mol\(^{-1}\))

Hint:

Remember, lowering of vapour pressure is related to the molality and the freezing point depression of the solution. For more precise calculations, consider the mole fraction of the solvent as well.

Answer 1

We can use the formula for freezing point depression to calculate the lowering in vapour pressure. The formula is: \[ \Delta T_f = \frac{K_f \cdot m}{1000} \] where: \(\Delta T_f\) is the freezing point depression, \(K_f\) is the cryoscopic constant (1.86 K kg mol\(^{-1}\) for water), and \(m\) is the molality of the solution. Since the solution freezes at -0.3°C, the freezing point depression \(\Delta T_f = 0.3\) K. Using the equation for freezing point depression, we can solve for \(m\), the molality of the solution. \[ m = \frac{\Delta T_f \cdot 1000}{K_f} = \frac{0.3 \cdot 1000}{1.86} = 161.29 \, \text{mol/kg} \] Next, the lowering of vapour pressure is given by Raoult's Law: \[ \frac{\Delta P}{P_0} = m \cdot X_1 \] where \(X_1\) is the mole fraction of the solvent (water) and \(P_0\) is the vapour pressure of pure water. Since the molality of the solution is high, we can estimate the lowering of vapour pressure by the formula: \[ \Delta P = P_0 \cdot \left( \frac{m}{1000} \right) \] Substituting the values: \[ \Delta P = 24.8 \cdot \left( \frac{161.29}{1000} \right) = 4.0 \, \text{mm Hg} \] Therefore, the lowering in vapour pressure is 4.0 mm Hg.