Multiple choice

Vapour pressure of $C_6H_6$ and $C_7H_8$ are 119 mm and 37 mm of Hg. Calculate molar composition of $C_6H_6$ and $C_7H_8$ in a mixture having V.P. of 80 mm of Hg. Also calculate vapour phase composition over the mixture:

  1. Liquid phase 52.44%, 47.56% vapour phase 78%, 22%

  2. Liquid phase 78%, 22% vapour phase 52.44%, 47.56%

  3. Liquid phase 60.45%, 39.50% vapour phase 82%, 18%

  4. Liquid phase 82%, 18% vapour phase 60.45%, 39.50%
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A Correct answer
Explanation

Using Raoult's Law: P_total = P_A*x_A + P_B*x_B. 80 = 119*x + 37*(1-x). 80 = 119x + 37 - 37x. 43 = 82x, so x = 43/82 = 0.5244 (52.44%). Then x_B = 47.56%. Vapour phase: y_A = P_A*x_A / P_total = 119*0.5244 / 80 = 0.78 (78%).

AI explanation

Using Raoult's law, the total vapor pressure of the liquid mixture is P = x1P1 + x2P2, so 80 = x1(119) + (1 - x1)(37). Solving this gives 82x1 = 43, so the mole fraction of C6H6 is 0.5244 (52.44 percent) and C7H8 is 0.4756 (47.56 percent). For the vapor phase, the mole fractions are y1 = x1P1/P_total = (0.5244)(119)/80 = 0.78 (78 percent) and y2 = 1 - 0.78 = 0.22 (22 percent).

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