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 $A = P^0_A$
Vapour pressure of $B = P^0_b$
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
$=250=P^{0}_{A}x_{A}+P^{0}_{B}x_{B}$
$250=\frac{1}{3}P^{0}_{A}+\frac{2}{3}P^{0}_{B} \,...\left(i\right)$
In 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
$=300=P^{0}_{A}x_{A}+P^{0}_{B}x_{B}
300=\frac{1}{2}P^{0}_{A}+\frac{1}{2}P^{0}_{B}\,...\left(ii\right)$
Multiplying equation $\left(i\right) by \frac{1}{2}$ and equation $\left(ii\right) by \frac{1}{3}$
$\frac{1}{6}P^{0}_{A}+\frac{2}{6}P^{0}_{A}=125$
$\frac{\frac{1}{6}P^{0}_{A}+\frac{1}{6}P^{0}_{B}=100}{\frac{1}{6}P^{0}_{B}=25}$
$P^{0}_{B}=25\times6=150\,mm \,Hg$
On substituting value of $P^{0}_{B}=25$ in equation (ii) we get
$=300=P^{0}_{A}\times\frac{1}{2}+150\times\frac{1}{2}$
$P^{0}_{A}=450\,mm\, Hg$