Mole Fraction of Component in Equilibrium Vapour Phase

To find the mole fraction of liquid component $\textbf{B}$ in the vapour phase ($y_B$) in equilibrium with the solution, we follow these steps:

Step 1: Calculate the mole fraction of components in the liquid phase

A $5\text{ molal}$ solution of $\textbf{B}$ in $\textbf{A}$ means that there are $5\text{ moles}$ of solute $\textbf{B}$ dissolved in $1\text{ kg}$ ($1000\text{ g}$) of solvent $\textbf{A}$.

• Moles of solute $\textbf{B}$, $n_B = 5\text{ mol}$
• Moles of solvent $\textbf{A}$, $n_A = \frac{\text{Mass of A}}{\text{Molar mass of A}} = \frac{1000\text{ g}}{50\text{ g mol}^{-1}} = 20\text{ mol}$

The mole fraction of $\textbf{A}$ in the liquid phase ($x_A$) is: $x_A = \frac{n_A}{n_A + n_B} = \frac{20}{20 + 5} = \frac{20}{25} = 0.8$

Similarly, the mole fraction of $\textbf{B}$ in the liquid phase ($x_B$) is: $x_B = \frac{n_B}{n_A + n_B} = \frac{5}{20 + 5} = \frac{5}{25} = 0.2$

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Step 2: Determine the partial vapour pressure of component A

According to Raoult's Law for an ideal solution, the partial vapour pressure of $\textbf{A}$ ($P_A$) is given by: $P_A = x_A \cdot P_A^\circ$

Given $P_A^\circ = 105\text{ mm Hg}$: $P_A = 0.8 \times 105\text{ mm Hg} = 84\text{ mm Hg}$

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Step 3: Determine the partial vapour pressure of component B

The total vapour pressure above the solution is $P_{\text{total}} = 100\text{ mm Hg}$. Using Dalton's Law of partial pressures: $P_{\text{total}} = P_A + P_B$

Substituting the known values: $100\text{ mm Hg} = 84\text{ mm Hg} + P_B$ $P_B = 100 - 84 = 16\text{ mm Hg}$

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Step 4: Calculate the mole fraction of B in the vapour phase

The mole fraction of $\textbf{B}$ in the vapour phase ($y_B$) is given by the ratio of its partial pressure to the total pressure: $y_B = \frac{P_B}{P_{\text{total}}} = \frac{16\text{ mm Hg}}{100\text{ mm Hg}} = 0.16$

Thus, the mole fraction of $\textbf{B}$ in the vapour phase is 0.16.