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

The decrease in freezing point can be used to calculate the reduction in vapor pressure. The formula for freezing point depression is: \[\Delta T_f = \frac{K_f \cdot m}{1000}\]Here, \(\Delta T_f\) represents 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. Given that the solution freezes at -0.3°C, the freezing point depression \(\Delta T_f\) is 0.3 K. We can rearrange the freezing point depression equation to solve for \(m\), the molality:\[m = \frac{\Delta T_f \cdot 1000}{K_f} = \frac{0.3 \cdot 1000}{1.86} = 161.29 \, \text{mol/kg}\]Raoult's Law provides the formula for vapor pressure lowering:\[\frac{\Delta P}{P_0} = m \cdot X_1\]In this equation, \(X_1\) is the mole fraction of the solvent (water), and \(P_0\) is the vapor pressure of pure water. Due to the high molality of the solution, the vapor pressure lowering can be approximated by:\[\Delta P = P_0 \cdot \left( \frac{m}{1000} \right)\]Substituting the known values:\[\Delta P = 24.8 \cdot \left( \frac{161.29}{1000} \right) = 4.0 \, \text{mm Hg}\]Consequently, the vapor pressure lowering is determined to be 4.0 mm Hg.