Exams › JEE Advanced › Chemistry

A 5.2 molal aqueous solution of methanol (CH3OH) is given. Determine the mole fraction of methanol in this solution.

0.086
0.050
0.100
0.190

Correct answer: 0.086

Solution

With 5.2 mol methanol and 55.56 mol water (1 kg), the mole fraction is 5.2/60.76 approximately 0.086.

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 →