Question

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added... vapour pressure of B in pure state is ___ mm Hg. (Nearest integer)

Hint:

For ideal solutions, the total vapor pressure change is linear with mole fraction changes. Set up simultaneous linear equations.

Answer 1

Correct Answer: 200

According to Raoult’s law, \[ P_{\text{solution}} = x_A P_A^\circ + x_B P_B^\circ \] Case I:
Number of moles: $n_A = 3$, $n_B = 1$
\[ x_A = \frac{3}{4}, \quad x_B = \frac{1}{4} \] Given vapour pressure: \[ 500 = \frac{3}{4}P_A^\circ + \frac{1}{4}P_B^\circ \] \[ 2000 = 3P_A^\circ + P_B^\circ \qquad (1) \] Case II:
After adding 1 mol of A: $n_A = 4$, $n_B = 1$
\[ x_A = \frac{4}{5}, \quad x_B = \frac{1}{5} \] Given vapour pressure of solution: \[ 520 = \frac{4}{5}P_A^\circ + \frac{1}{5}P_B^\circ \] \[ 2600 = 4P_A^\circ + P_B^\circ \qquad (2) \] Subtracting equation (1) from equation (2): \[ 600 = P_A^\circ \] Substituting $P_A^\circ = 600$ in equation (1): \[ 2000 = 3(600) + P_B^\circ \] \[ P_B^\circ = 200 \text{ mm Hg} \] \[ \therefore \text{Vapour pressure of B in pure state} = 200 \text{ mm Hg} \]