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= 3 4 M _ Y . 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 v

Step-by-step solution

The relationship between molar masses of the two solvents is M _ X = 3 4 M _ Y The relative lowering of vapour pressure of the two solutions is ( P P )_ X = m ( P P )_ Y But, the relative lowering 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 1 ~kg ( or 1000 ~g ) of solvent. The number of moles of solvent = 1000 ~g M The mole fraction of solute = 5 1000 / M = M 5 1000 hence M _ X 5 1000 = m M _ Y 5 1000 . Sub