Exams › JEE Advanced › Chemistry

What is the vapour pressure of solution M?

1. 39.3 mm Hg
2. 36.0 mm Hg
3. 29.5 mm Hg
4. 28.8 mm Hg

Correct answer: 29.5 mm Hg

Solution

For an ideal solution, the vapour pressure of the solution is given by Raoult's law: $P = x_{\text{solvent}}P^0$. Using the composition of solution M from the given data, the calculated vapour pressure comes out to 29.5 mm Hg. Hence, that is the correct option.

Related JEE Advanced Chemistry questions

• Two liquids P and Q form an ideal solution. At 300 K, the vapor pressure of a solution containing 1 mole of P and 3 moles of Q is 550 mmHg. If 1 mole of Q is added to this mixture at the same temperature, the vapor pressure increases by 10 mmHg. What are the vapor pressures of pure P and Q, in mmHg?
• Two liquids A and B form an ideal mixture. At 30°C, a solution containing 1 mole of A and 2 moles of B has a total vapor pressure of 250 mmHg. When 1 additional mole of A is added to this mixture, the vapor pressure rises to 300 mmHg. What are the vapor pressures of pure A and B at 30°C?
• What is the freezing temperature of solution M?
• What is the approximate value of the van’t Hoff factor in this case?
• A compound MX$_2$ dissociates into M$^{2+}$ and X$^-$ ions in water, with degree of dissociation $\alpha = 0.5$. What is the ratio of the actual freezing point depression of the solution to the freezing point depression if no dissociation occurred?
• On dissolving 0.5 g of a non-volatile, non-ionic solute in 39 g of benzene, the vapour pressure decreases from 650 mm Hg to 640 mm Hg. The depression of freezing point of benzene (in K) upon addition of the solute is ____. (Given data: molar mass of benzene = 78 g mol$^{-1}$ and molal freezing point depression constant of benzene = 5.12 K kg mol$^{-1}$)

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →