a
The relationship between molar masses of the two solvents is

\(M_X\,=\,\frac {3}{4}M_Y\)    .... \((i)\)

The relative lowering of vapour pressure of two solution is

\({\left( {\frac{{\Delta P}}{P}} \right)_X}\, = \,m{\left( {\frac{{\Delta P}}{P}} \right)_Y}\)

But, the relative loering of vaoour pressure of solutin is directly proportional to the mole fraction of solution.

Given \(5\,molal\) solution , mans \(5\,molal\) of solute are dissolved in \(1\,kg\) (or \(1000\,g\)) of solvent.

The number of moles of solvent \(=\,\frac {1000\,g}{M}\)

The mole fraction of solution \(=\,\frac {5}{1000/M}\)

                                              \(=M\times \,\frac {5}{1000}\)

hence \(M_X\times \frac {5}{1000}\) \(=\,m\times M_Y\times \frac {5}{1000}\)   .... \((ii)\)

Substitute equation \((i)\) in equation \((ii)\)

\(\frac {3}{4}\times M_Y\times \frac {5}{1000}\) \(=\,m\times M_Y\times \frac {5}{1000}\)

\(m=\frac {3}{4}\)