Question

Vapour pressure of pure A is $\left(P_A^{\circ }\right)=100$ mm Hg. Vapour pressure of pure B is $\left(P_B^{\circ }\right)=150$ 150mm Hg. If 2 moles of A and 3 moles of B are mixed to form ideal solution, the vapour pressure of solution will be

• A. 135 mm
• B. 135 mm
• C. 140 mm
• D. 145 mm

Answer 1

Answer: (B)

Step 1:
We know that definition of mole fraction =$\frac{No.ofmolesofcomponent}{Totalno.ofmolesofallcomponents}$
Mole fraction of $A\left(x_A\right)=\frac{2}{2+3}=\frac{2}{5}$
Mole fraction of $A\left(x_B\right)=\frac{3}{2+3}=\frac{3}{5}$

Step 2:
$P_T=P_A^{\circ }\times X_A+P^{\circ }B\times X_B$
$=100\times \frac{2}{5}+150\times \frac{3}{5}=130mmHg$

Hence, option (B) is correct.