Question

Liquids A and B form an ideal solution at 30°C, the total vapour pressure of a solution containing 1 mol of A and 2 mol of B is 250 mmHg. The total vapour pressure becomes 300 mmHg when 1 more mol of A is added to the first solution. The vapour pressures of pure A and B at the same temperature are

• A. 150, 450 mmHg
• B. 125, 150 mmHg
• C. 450, 150 mmHg
• D. 250, 300 mmHg

Answer 1

Answer: (C)

450, 150 mmHg

Answer 2

Let vapour pressure of $A = P^0_A$ and vapour pressure of $B = P^0_b$

In the first case,

Mole fraction of $A\left(x_{A}\right)=\frac{1}{1+2}=\frac{1}{3}$

Mole fraction of $B\left(x_{B}\right)=\frac{2}{1+2}=\frac{2}{3}$

According to Raoult's law,

Total vapour pressure $=250=P^{0}_{A}x_{A}+P^{0}_{B}x_{B}$

$250=\frac{1}{3}P^{0}_{A}+\frac{2}{3}P^{0}_{B} \,...\left(i\right)$

In the second case,

Mole fraction of $A\left(x_{A}\right)=\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}$

Mole fraction of $B\left(x_{B}\right)=\frac{2}{4}=\frac{1}{2}$

$\therefore$ Total vapour pressure

$300=P^{0}_{A}x_{A}+P^{0}_{B}x_{B}$

$300=\frac{1}{2}P^{0}_{A}+\frac{1}{2}P^{0}_{B}...(ii)$

Multiplying equation $(i)$ by $\frac{1}{2}$ and equation $\left(ii\right)$ by $\frac{1}{3}$

$\frac{1}{6}P^{0}_{A}+\frac{2}{6}P^{0}_{A}=125$

$\frac{\frac{1}{6}P^{0}_{A}+\frac{1}{6}P^{0}_{B}=100}{\frac{1}{6}P^{0}_{B}=25}$

$P^{0}_{B}=25\times6=150\,mm \,Hg$

Now, substitute the value of $P^{0}_{B}=25$ in equation (ii), we get

$300=P^{0}_{A}\times\frac{1}{2}+150\times\frac{1}{2}$

$P^{0}_{A}=450\,mmHg$