Solutions and Colligative PropertieshardMCQ SINGLE

A liquid mixture of ‘A’ and ‘B’ (assume ideal solution) is placed in a cylinder and piston arrangemeSolutions and Colligative Properties Chemistry Question

Question

A liquid mixture of ‘A’ and ‘B’ (assume ideal solution) is placed in a cylinder and piston arrangement. The piston is slowly pulled out isothermally so that the volume of liquid decreases and that of the vapour increases. At the instant when the quantity of the liquid still remaining is negligibly small, the mole fraction of ‘A’ in the vapour is 0.4. If $P_A^\circ = 0.4 \text{ atm}, P_B^\circ = 1.2 \text{ atm}$ at this temperature, the total pressure at which the liquid has almost evaporated, is

Answer: A

💡 Solution & Explanation

When the last drop of liquid remains, the composition of the entire vapour phase equals the overall initial composition. So $Y_A = 0.4, Y_B = 0.6$. The total pressure $P$ relates to vapour composition by $1/P = Y_A / P_A^\circ + Y_B / P_B^\circ = 0.4 / 0.4 + 0.6 / 1.2 = 1 + 0.5 = 1.5$. Therefore, $P = 1 / 1.5 = 0.667 \text{ atm}$. Therefore, correct answer is A.

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Advanced Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry