Multiple choice

A mixture of two miscible liquids $A$ and $B$ is distilled under equilibrium conditions at $1$ atm pressure. The mole fraction of $A$ in solution and vapour phase are $0.30$ and $0.60$ respectively. Assuming ideal behaviour of the solution and the vapour, calculate the ratio of the vapour pressure of pure $A$ to that of pure $B$.

  1. $4.0$
  2. $3.5$
  3. $2.5$
  4. $1.85$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Raoult's Law: P_A = x_A * P_A_pure, P_B = x_B * P_B_pure. In vapor phase, y_A = P_A / P_total. y_A = 0.6, x_A = 0.3, x_B = 0.7. P_A = 0.3 * P_A_pure, P_B = 0.7 * P_B_pure. P_total = 0.3*P_A_pure + 0.7*P_B_pure. y_A = (0.3*P_A_pure) / (0.3*P_A_pure + 0.7*P_B_pure) = 0.6. 0.3*P_A_pure = 0.18*P_A_pure + 0.42*P_B_pure. 0.12*P_A_pure = 0.42*P_B_pure. Ratio = P_A_pure / P_B_pure = 0.42 / 0.12 = 3.5.

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