The relationship between molar masses of the two solvents is MX=
3
4
MY... (i) The relative lowering of vapour pressure of the two solutions is (
∆P
P
)X=m(
∆P
P
)Y But, the relativelowering of vapour pressure of solutions is directly proportional to the mole fraction of solute. Given 5 molal solution, means 5 moles of solute are dissolved in 1kg (or 1000g ) of solvent. The number of moles of solvent =
1000g
M
The mole fraction of solute =
5
1000∕M
=M×
5
1000
Hence, MX×
5
1000
=m×MY×
5
1000
... (ii) Substitute equation (i) in equation (ii)