Question

At \(298\ K\), vapour pressures of two pure liquids A and B are \(200\) and \(400\ \text{mm Hg}\) respectively. If mole fractions of A and B in solution are \(0.7\) and \(0.3\) respectively, what is the mole fraction of B in vapour phase?

• A. \(0.279\)
• B. \(0.721\)
• C. \(0.538\)
• D. \(0.462\)

Hint:

For an ideal binary liquid solution, use Raoult's law: \[ p_i=x_ip_i^\circ \] and mole fraction in vapour phase: \[ y_i=\frac{p_i}{p_{\text{total}}} \]

Answer 1

The Correct Option is D

Step 1: Write Raoult's law.
For an ideal solution, \[ p_A=x_Ap_A^\circ \] and \[ p_B=x_Bp_B^\circ \] Given, \[ p_A^\circ=200\ \text{mm Hg} \] \[ p_B^\circ=400\ \text{mm Hg} \] \[ x_A=0.7,\quad x_B=0.3 \]

Step 2: Calculate partial pressure of A.
\[ p_A=x_Ap_A^\circ \] \[ p_A=0.7\times200 \] \[ p_A=140\ \text{mm Hg} \]

Step 3: Calculate partial pressure of B.
\[ p_B=x_Bp_B^\circ \] \[ p_B=0.3\times400 \] \[ p_B=120\ \text{mm Hg} \]

Step 4: Calculate total vapour pressure.
\[ p_{\text{total}}=p_A+p_B \] \[ p_{\text{total}}=140+120 \] \[ p_{\text{total}}=260\ \text{mm Hg} \]

Step 5: Calculate mole fraction of B in vapour phase.
Mole fraction of B in vapour phase is \[ y_B=\frac{p_B}{p_{\text{total}}} \] \[ y_B=\frac{120}{260} \] \[ y_B=0.462 \]

Step 6: Final conclusion.
Therefore, the mole fraction of B in vapour phase is \[ \boxed{0.462} \] Hence, the correct option is \[ \boxed{(4)} \]