Question

At 300 k the vapour pressure of an ideal solution containing 1 mole of liquid a and 2 mol of liquid b is 500 mm of hg the vapour pressure of the solution increases by 25 mm of hg if one mote mole of b is added to the above udeal solution at 300 k thwn the vapour pressure of a in its pute state is

Answer 1

Answer: The vapour pressure of A in its pure state is 290 mmHg.

Explanation: According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2P_2^0$

where, x = mole fraction

$p^0$ = pressure in the pure state

$x_A=\frac{1}{1+2}=\frac{1}{3}=0.33$

$x_B=\frac{2}{1+2}=\frac{2}{3}=0.67$

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$

$p_{total}=x_Ap_A^0+x_BP_B^0$

$500=0.33\times p_A^0+0.67\times p_B^0$     (1)

Also: when 1 mole of B is added to the above solution

$x_A=\frac{1}{1+3}=\frac{1}{4}=0.25$

$x_B=\frac{3}{1+3}=\frac{3}{4}=0.75$

$525=0.25\times p_A^0+0.75\times p_B^0$         (2)

Solving 1 and 2

$p_A^0=290mmHg$