The vapour pressure of pure benzene at 88°C is 960 mm and that of toluene at the same temperature is 380 mm of benzene (mm Hg). At what mole fraction of benzene, the mixture will boil at 88°C?

Question

Prashant Kotian · Accepted Answer

Correct answer: A. The mixture will boil when its total vapour pressure equals atmospheric pressure (1 atm = 760 mm Hg). Using Raoult's law: P_total = P_benzene^o * X_benzene + P_toluene^o * (1 - X_benzene) => 760 = 960 * X_benzene + 380 * (1 - X_benzene) => 760 = 580 * X_benzene + 380 => 380 = 580 * X_benzene => X_benzene = 380 / 580 ≈ 0.655.

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