An organic liquid is purified by steam distillation. The distillation is carried out at a temperature where the total atmospheric pressure is 760 mm Hg. The vapour pressure of water at this distillation temperature is 700 mm Hg. If the distillate contains 1.5 g of the organic liquid and 1.8 g of water, the molar mass of the organic liquid is: (Given: Molar mass of water is 18 g mol ⁻¹ )

Step-by-step solution

In steam distillation, the total pressure is the sum of the partial vapour pressures of the organic liquid ( P_l ) and water ( P_w ). P_ total = P_l + P_w 760 = P_l + 700 P_l = 60 mm Hg The ratio of the masses of the organic liquid ( w_l ) and water ( w_w ) in the distillate is related to their partial pressures and molar masses by the equation: w_l w_w = P_l M_l P_w M_w where M_l and M_w are the molar masses of the organic liquid and water, respectively. Substituting the given values: 1.5 1.8 = 60 M_l 700 18 5 6 =