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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.

Options

  1. A1 g
  2. B1.27 g
  3. C0.787 g
  4. D13.43 g

Correct answer

B. 1.27 g

Step-by-step solution

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

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