Question

Liquids ‘X’ and ‘Y’ form an ideal solution. The vapour pressure of pure ‘X’ and pure ‘Y’ are 120 mm Hg and 160 mm Hg respectively. Calculate the vapour pressure of the solution containing equal moles of ‘X’ and ‘Y’.

Hint:

Raoult's law can be used for ideal solutions to determine the total vapour pressure by summing the partial pressures based on the mole fractions of the components.

Answer 1

For an ideal solution, the vapour pressure of the solution can be calculated using Raoult's law, which states that the vapour pressure of each component is proportional to its mole fraction in the solution. The total vapour pressure is the sum of the partial pressures of the components:
\[ P_{\text{total}} = P_{\text{X}} + P_{\text{Y}} \] The partial pressures are calculated as: \[ P_{\text{X}} = P_{\text{X}}^0 \times x_{\text{X}}, \quad P_{\text{Y}} = P_{\text{Y}}^0 \times x_{\text{Y}} \] where \(P_{\text{X}}^0\) and \(P_{\text{Y}}^0\) are the vapour pressures of pure components and \(x_{\text{X}}\) and \(x_{\text{Y}}\) are the mole fractions of X and Y. Since equal moles are taken, the mole fraction of each component is 0.5:
\[ P_{\text{X}} = 120 \times 0.5 = 60 \text{ mm Hg}, \quad P_{\text{Y}} = 160 \times 0.5 = 80 \text{ mm Hg} \] Thus, the total vapour pressure of the solution is: \[ P_{\text{total}} = 60 + 80 = 140 \text{ mm Hg} \]