Question

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 are \(M_x\) and \(M_y\), respectively where \(M_x = \frac{3}{4}M_y\). The relative lowering of vapor 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 the solvent, the value of "m" is :

• A. \(\frac{4}{3}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{1}{4}\)

Hint:

For a dilute solution, relative lowering of vapour pressure \(\approx n_2/n_1\), which is proportional to molality times the solvent molar mass.

Answer 1

The Correct Option is B

Relative lowering of vapour pressure: \(\dfrac{\Delta P}{P^0} = x_2 = \dfrac{n_2}{n_1+n_2} \approx \dfrac{n_2}{n_1}\) (dilute solution).
For 1 kg of solvent: \(n_2 = 5\) mol (5 molal) and \(n_1 = \dfrac{1000}{M}\).
So \(\dfrac{\Delta P}{P^0} = \dfrac{5M}{1000}\), i.e. proportional to the solvent molar mass.
\(m = \dfrac{(\Delta P/P^0)_X}{(\Delta P/P^0)_Y} = \dfrac{M_x}{M_y} = \dfrac{3}{4}\).
Answer: (2) \(\frac{3}{4}\)