Question

At ${80}^{\circ }$C the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liq ‘B’ is 1000 mm Hg. If a mixture of both the liquids form an ideal solution which boils at ${80}^{\circ }C$ and 1 atm pressure, the mole fraction of ‘A’ in the liquid mixture is:

• A. 0.4
• B. 0.2
• C. 0.5
• D. 0.8

Answer 1

The correct option is C 0.5

$P_A^{\circ }{X}_A+P_B^{\circ }{X}_B=P_{mix}\left(520\right)X_A+\left(1000\right)(1-X_A)=760520X_A-1000X_A=760-1000-480X_A=-240\therefore X_A=\frac{1}{2}=0.5$