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 to this solution, vapour pressure of the solution increases by 20 mm Hg. Vapour pressure (in mm Hg) of B in pure state is ____

Two liquids A and B form an ideal solution vapour pressure numerical

Q. 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 to this solution, vapour pressure of the solution increases by 20 mm Hg.

Vapour pressure (in mm Hg) of B in pure state is _____. (Nearest integer)

Correct Answer: 200

Explanation

Since the solution is ideal, Raoult’s law is applicable.

Let vapour pressures of pure components be:

$$ P_A^\circ = P_A,\quad P_B^\circ = P_B $$

Initial solution:

Moles of A = 3, moles of B = 1

Total moles = 4

Mole fractions:

$$ x_A = \frac{3}{4},\quad x_B = \frac{1}{4} $$

Total vapour pressure:

$$ P = x_A P_A + x_B P_B $$

$$ \frac{3}{4}P_A + \frac{1}{4}P_B = 500 $$

Multiplying by 4:

$$ 3P_A + P_B = 2000 \quad \text{(Equation 1)} $$

After adding 1 mol of A:

Moles of A = 4, moles of B = 1

Total moles = 5

New mole fractions:

$$ x_A = \frac{4}{5},\quad x_B = \frac{1}{5} $$

New vapour pressure = 500 + 20 = 520 mm Hg

Applying Raoult’s law again:

$$ \frac{4}{5}P_A + \frac{1}{5}P_B = 520 $$

Multiplying by 5:

$$ 4P_A + P_B = 2600 \quad \text{(Equation 2)} $$

Subtract Equation (1) from Equation (2):

$$ (4P_A + P_B) - (3P_A + P_B) = 2600 - 2000 $$

$$ P_A = 600 $$

Substitute $P_A = 600$ in Equation (1):

$$ 3(600) + P_B = 2000 $$

$$ P_B = 200 $$

Therefore, vapour pressure of B in pure state is 200 mm Hg.

Explanation

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