A liquid is immiscible in water was steam distilled at 95.2 o C at a pressure of 0.983 atm. What is the mass of the liquid present per gram of water in the distillate. Molar mass of the liquid is 134.3 g/mol and the vapour pressure of water is 0.84 atm. Also, Vapour pressure of pure liquid is 0.143 atm.

Question

A liquid is immiscible in water was steam distilled at 95.2 o C at a pressure of 0.983 atm. What is the mass of the liquid present per gram of water in the distillate. Molar mass of the liquid is 134.3 g/mol and the vapour pressure of water is 0.84 atm. Also, Vapour pressure of pure liquid is 0.143 atm., 1 g, 1.27 g, 0.787 g, 13.43 g

MARKS App · Accepted Answer

p T = 0.983 atm p water ∘ = 0.84 atm ∴ p liquid ∘ = 0.143 atm Apply Dalton's law of vapour phase p o water p o liquid = 0.84 0.143 = n H 2 O n L i q u i d 0.84 0.143 = 1 18 w e 134.3 = 134.3 18 × w e w e = 134.3 18 × 0.143 0.84 = 1.27 g

Search any question & find its solution

Looking for more such questions to practice?