Exams › JEE Main › Chemistry

Two 5 molal solutions are prepared by dissolving a non-electrolyte, non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents Mx and My, respectively where Mx = 3/4 My. The relative lowering of vapour pressure of the solution in X is "m" times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of solvent, the value of "m" is - [JEE-Main On line-2018]

1. 3/4
2. 1/2
3. 1/4
4. 4/3

Correct answer: 3/4

Solution

For dilute solutions RLVP ~ n_solute/n_solvent. At equal molality, moles solvent per kg = 1000/M, so RLVP is proportional to M. Thus m = Mx/My = 3/4.

Related JEE Main Chemistry questions

• A sulphuric acid solution has a concentration of 3.60 M and contains 29% H2SO4 by mass. If the molar mass of H2SO4 is 98 g mol⁻¹, what is the density of the solution in g mL⁻¹?
• Which one of the following substances is absent in clear hard water?
• Fractional distillation is chosen in a situation where
• Car radiators use an aqueous ethylene glycol solution as antifreeze. How much ethylene glycol should be mixed with 5 kg of water so that the freezing point is lowered to -0.3 °C? (K_f = 1.86 K kg mol⁻¹)
• A solution contains a non-volatile solute of molecular mass M₂. Which expression can be used to determine the molecular mass of the solute using osmotic pressure?
• At 298 K, the Henry’s law constant for oxygen is 1.4 × 10⁻³ mol L⁻¹ atm⁻¹. If oxygen is at a partial pressure of 0.5 atm, what mass of oxygen will dissolve in 100 mL of water at 298 K?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →