At , the vapour pressure of pure liquid A is and that of pure liquid B is . If a mixture of A and B — Solutions and Colligative Properties Chemistry Question
Question
At $80^\circ\text{C}$, the vapour pressure of pure liquid A is $520\text{ mm Hg}$ and that of pure liquid B is $1000\text{ mm Hg}$. If a mixture of A and B boils precisely at $80^\circ\text{C}$ under an external pressure of $1\text{ atm}$, what is the mole percentage of A in the liquid mixture?
Answer: B
💡 Solution & Explanation
Boiling occurs when $P_{total} = P_{ext} = 1\text{ atm} = 760\text{ mm Hg}$. Thus, $760 = P^0_A X_A + P^0_B (1 - X_A) \implies 760 = 520 X_A + 1000(1 - X_A)$. Solving this gives $760 = 1000 - 480 X_A \implies 480 X_A = 240 \implies X_A = 0.5$. The mole percentage is $50\%$.
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