The expression to compute molar mass of a solute from the relative lowering of vapour pressure of a solvent is where the various symbols have their usual meanings. 

The relative reduction of vapour pressure is expressed as follows: $\frac{\mathrm{\Delta P}}{{\mathrm{P}}^0}=\frac{{\mathrm{P}}^0-\mathrm{P}}{{\mathrm{P}}^0}={\mathrm{X}}_2$.....(1)

Here, $\frac{\mathrm{\Delta P}}{{\mathrm{P}}^0}$ and $\frac{{\mathrm{P}}^0-\mathrm{P}}{{\mathrm{P}}^0}$ denotes a relative decrease in vapour pressure, while X₂ denotes the solute's molecular weight.

But the solute's mole fraction

${\mathrm{X}}_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}=\frac{\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}}{\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}+\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1}}$ .....(2)

Here, W₁ and M₁ represent the mass and molar mass of the solvent, while W₂ and M₂ represent the mass and molar mass of the solute. The solute and solvent's respective molecular weights are n₁ and n₂, respectively.

for weak solutions ${\mathrm{n}}_1>>{\mathrm{n}}_2$

Therefore, n₂ can be disregarded in favour of n₁.
Equation (2) consequently becomes

${\mathrm{X}}_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}=\frac{\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}}{\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1}}$  ......(3)

the results of equations (1) and (3)

$\frac{\mathrm{\Delta P}}{{\mathrm{P}}^0}=\frac{{\mathrm{P}}^0-\mathrm{P}}{{\mathrm{P}}^0}=\frac{\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}}{\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1}}$

Here, to compute molar mass of a solute from the relative lowering of vapour pressure of a solvent is where the various symbols have their usual meanings $\begin{array}{r}{M}_2=m_2/\left\{\left(m_1/M_1\right)\left(-\mathrm{\Delta }{p}_1/p_1^{\ast }\right)\right\}\end{array}$.