Question

At 300 K, the vapour pressure of an ideal solution containing 3 mol of A and 2 mol of B is 600 torr. At the same temperature, if 1.5 mol of A and 0.5 mol of C (non-volatile) are added to this solution, the vapour pressure of solution increases by 30 torr. What is the value of $P_B^0$ (in torr)?

• A. 940
• B. 405
• C. 90
• D. None of these

Answer 1

The correct option is C 90

According to Raoult's law, the total vapour pressure of solution is given by
$P$ $=$ $P_A^0$${\chi }_A$ $+$ $P_B^0$${\chi }_B$

Initial ${\chi }_A=\frac{3}{3+2}$ and ${\chi }_B=\frac{2}{3+2}$

$600$$=$$P_A^0$($\frac{3}{3+2}$) $+$ $P_B^0$($\frac{2}{2+3}$);

$3$$P_A^0$ $+$ $2$$P_B^0$ $=$$3000\to eq.\left(1\right)$

After the additon of 1.5 mol of A and 0.5 mol of a non-volatile solute C, new vapour pressure is given by:

$630=P_A^0$($\frac{4.5}{4.5+2+0.5})+P_B^0$($\frac{2}{4.5+2+0.5}$)

$4.5$$P_A^0$ $+$ $2$$P_B^0$ $=$ $4410\to eq.\left(2\right)$

On subtracting eq. 1 from 2, we obtain
$1.5$$P_A^0$ $=$ $1410$
So $P_A^0$ = 940 torr
Substituting this value in equation 1 we get:
$P_B^0$$=$ $90$ torr