The immiscible liquid system containing aniline-water boils at 98°C under a pressure of 760 mm. At t — Solutions and Colligative Properties Chemistry Question
Question
The immiscible liquid system containing aniline-water boils at 98°C under a pressure of 760 mm. At this temperature, the vapour pressure of water is 700 mm. If aniline is distilled in steam at 98°C, what per cent of total weight of the distillate will be aniline?
💡 Solution & Explanation
For immiscible liquids, the total vapour pressure is: P_total = P_water^o + P_aniline^o => 760 = 700 + P_aniline^o => P_aniline^o = 60 mm. The mass ratio in the distillate is: w_aniline / w_water = (P_aniline^o * M_aniline) / (P_water^o * M_water). Given M_aniline (C6H5NH2) = 93 g/mol, M_water = 18 g/mol: w_aniline / w_water = (60 * 93) / (700 * 18) = 5580 / 12600 ≈ 0.4428. Percentage of aniline = w_aniline / (w_aniline + w_water) * 100 = 0.4428 / (1 + 0.4428) * 100 ≈ 30.7%.