Exams › JEE Advanced › Chemistry

Two aqueous solutions S1 and S2 are separated by a semipermeable membrane. The vapour pressure of S1 is higher than that of S2. What will happen?

1. More solvent will flow from S1 to S2
2. More solvent will flow from S2 to S1
3. Solvent flows at equal rates in both directions
4. There is no flow of solvent

Correct answer: More solvent will flow from S1 to S2

Solution

Vapour pressure of a solution is inversely related to solute concentration (Raoult's law: VP = VP_pure * x_solvent, so higher VP means higher x_solvent = lower solute concentration = lower osmolarity). Higher VP for S1 means S1 has lower solute concentration and lower osmotic pressure compared to S2. Water (solvent) flows by osmosis from the region of lower osmotic pressure (S1) to higher osmotic pressure (S2), i.e., from S1 to S2.

Related JEE Advanced Chemistry questions

• Two liquids P and Q create an ideal solution. At 300 K, the vapor pressure of a solution with 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 rises 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 with 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 pure B at 30°C?
• A compound M X₂ separates into M²⁺ and X⁻ ions in water, with a dissociation extent (α) of 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 to 39 g of benzene, its vapor 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 and the molal freezing point depression constant of benzene are 78 g mol⁻¹ and 5.12 K kg mol⁻¹, respectively)
• At 25°C, liquids A and B create an ideal solution for all proportions of A and B. Two mixtures with mole fractions of A as 0.25 and 0.50 exhibit total vapor pressures of 0.3 bar and 0.4 bar, respectively. Determine the vapor pressure of pure liquid B in bar.
• The ratio of solubilities of N2 and O2 gases in water from air at 25 degrees C and 1 atm total pressure is (given that air contains 80% N2 and 20% O2 by volume, and Henry's law constants K_H(N2) = 8.42 * 10⁴ atm and K_H(O2) = 4.46 * 10⁴ atm in terms of mole fraction):

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →