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 mm Hg. The total vapour pressure becomes 300 mm Hg 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 mm Hg
• B. 125,150 mm Hg
• C. 450,150 mm Hg
• D. 250,300 mm Hg

Answer 1

Answer: (C)

Let vapour pressure of $A=P_A^0$
Vapour pressure of $B=P_b^0$
In first solution,
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
$=250=P_A^0{x}_A+P_B^0{x}_B$
$250=\frac{1}{3}{P}_A^0+\frac{2}{3}{P}_B^0...\left(i\right)$
In second solution
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
$=300=P_A^0{x}_A+P_B^0{x}_B<br/>300=\frac{1}{2}{P}_A^0+\frac{1}{2}{P}_B^0...\left(ii\right)$
Multiplying equation $\left(i\right)by\frac{1}{2}$ and equation $\left(ii\right)by\frac{1}{3}$
$\frac{1}{6}{P}_A^0+\frac{2}{6}{P}_A^0=125$
$\frac{\frac{1}{6}{P}_A^0+\frac{1}{6}{P}_B^0=100}{\frac{1}{6}{P}_B^0=25}$
$P_B^0=25\times 6=150mmHg$
On substituting value of $P_B^0=25$ in equation (ii) we get
$=300=P_A^0\times \frac{1}{2}+150\times \frac{1}{2}$
$P_A^0=450mmHg$