Passage: Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase. Give

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Passage: Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase. Given: The gas constant R = 0.08 L atm K^ -1 mol^ -1 Molar mass of A is 50 g mol^ -1 Molar mass of B is 57 g mol^ -1 Density of liquid B at 300 K is 0.5 g/mL 1 atm = 760 mm Hg Question: The mole fraction of B in vapour phase which is in equilibrium with this solution is ____.

Quantrex Academy · Accepted Answer

Correct answer: A. A

Molality of B in A is 5 m, which means 5 moles of B are dissolved in 1 kg (1000 g) of solvent A. Number of moles of A, n_A = 1000 50 = 20 mol Mole fraction of A in the liquid phase, x_A = n_A n_A + n_B = 20 20 + 5 = 0.8 Mole fraction of B in the liquid phase, x_B = 1 - 0.8 = 0.2 According to Raoult's law, the partial vapour pressure of A is: P_A = P_A^0 x_A = 105 0.8 = 84 mm Hg Given total vapour pressure, P_T = 100 mm Hg Partial vapour pressure of B, P_B = P_T - P_A = 100 - 84 = 16 mm Hg Mole fraction of B in the vapour phase, y_B = P_B P_T = 16 100 = 0.16 Answer: 0.16

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