Question

An ideal solution is prepared by mixing 3 moles of liquid B and 7 moles of liquid A. If the vapour pressure of pure liquid A is 150 mm Hg and pure liquid B is 140 mm Hg. Calculate the ratio of number of moles of A (in vapour phase) to the number of moles of B (in vapour phase).

Answer 1

2.5

Answer 2

1. Calculate Mole Fractions in Liquid Phase:
   The total moles of liquid are $n_A + n_B = 7 + 3 = 10$ moles.
   Mole fraction of A in liquid ($X_A$) = $\frac{n_A}{n_{total}} = \frac{7}{10} = 0.7$.
   Mole fraction of B in liquid ($X_B$) = $\frac{n_B}{n_{total}} = \frac{3}{10} = 0.3$.

2. Calculate Partial Pressures in Vapor Phase (Raoult's Law):
   The partial pressure of a component in the vapor phase is given by Raoult's Law: $P_i = X_i P_i^0$.
   Partial pressure of A ($P_A$) = $X_A \cdot P_A^0 = 0.7 \times 150 \text{ mm Hg} = 105 \text{ mm Hg}$.
   Partial pressure of B ($P_B$) = $X_B \cdot P_B^0 = 0.3 \times 140 \text{ mm Hg} = 42 \text{ mm Hg}$.

3. Calculate Ratio of Moles in Vapor Phase (Dalton's Law):
   According to Dalton's Law of Partial Pressures, the mole fraction of a component in the vapor phase ($Y_i$) is equal to its partial pressure divided by the total vapor pressure ($P_{total}$).
   $Y_A = \frac{P_A}{P_{total}}$ and $Y_B = \frac{P_B}{P_{total}}$.
   The ratio of the number of moles of A to B in the vapor phase is equal to the ratio of their mole fractions in the vapor phase, which simplifies to the ratio of their partial pressures:
   $\frac{n_A(\text{vapor})}{n_B(\text{vapor})} = \frac{Y_A}{Y_B} = \frac{P_A/P_{total}}{P_B/P_{total}} = \frac{P_A}{P_B}$.
   Ratio = $\frac{105 \text{ mm Hg}}{42 \text{ mm Hg}}$.
   Simplifying the ratio by dividing both numerator and denominator by 21:
   Ratio = $\frac{105 \div 21}{42 \div 21} = \frac{5}{2} = 2.5$.