A mixture of benzene and toluene contains 30% by mass of toluene. At 30 °C, vapour pressure of pure toluene is 36.7 mmHg and that of pure benzene is 118.2 mmHg. Assuming that the two liquids form ideal solution, calculate partial pressure of each con…

Answer: \(p_{\text{toluene}} \approx 9.78\) mmHg, \(p_{\text{benzene}} \approx 86.7\) mmHg

Take 100 g of mixture: toluene (C₇H₈, M = 92 g mol^–1) = 30 g; benzene (C₆H₆, M = 78 g mol^–1) = 70 g.

\(n_{T} = \dfrac{30}{92} = 0.3261\) mol, \(n_{B} = \dfrac{70}{78} = 0.8974\) mol, total = 1.2235 mol

\(x_{T} = \dfrac{0.3261}{1.2235} = 0.2665\), \(x_{B} = 1 - 0.2665 = 0.7335\)

By Raoult’s law:

\(p_{T} = x_{T}\,p^0_{T} = 0.2665 \times 36.7 = 9.78\) mmHg

\(p_{B} = x_{B}\,p^0_{B} = 0.7335 \times 118.2 = 86.70\) mmHg

(Total vapour pressure = 96.48 mmHg.)