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 MX and MY, respectively where MX = 34MY. 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?

The relationship between molar masses of two solvents is MX = 34MY. (1)
The relative lowering of vapour pressure of two solutions is (ΔP/P)X = m(ΔP/P)Y
But, the relative lowering of vapour pressure of solution is directly proportional to the mole fraction of solute.
Mx × 5/1000 = m × MY × 5/1000 (2)
Substitute equation (1) in equation (2).
34 × MY × 5/1000 = m × MY × 5/1000
m = 34
Note: 5 molal solution means 5 moles of solute are dissolved in 1 kg (or 1000 g) of solvent. The number of moles of solvent = 1000g/M
The mole fraction of solute = 5/(1000/M + 5) ≈ 5/(1000/M) = M × 5/1000