Question

Two liquids A and B form an ideal solution. The vapour pressure of pure A and pure B are 66mm of Hg and 88mm of Hg, respectively. Calculate the composition of vapour A in the solution which is equilibrium and whose molar volume is 36%.

• A. 0.43
• B. 0.7
• C. 0.3
• D. 0.5

Answer 1

The correct option is A 0.43

$P_t=P_A+P_B=66mm+88mm=154mm$ of $Hg$

Now, $P_A$ $=X_A\times P_T$

Therefore, $X_A$ =$\frac{P_A}{P_T}=\frac{66}{154}$ $=0.43$

Therefore, composition of Vapour A is $0.43$.