What is relative lowering of vapour pressure? How is it useful to determine the molar mass of a solute?

i) The ratio of lowering of vapour pressure (p0 – p_S) to the vapour pressure of the pure
solvent (p0) is known as the relative lowering of vapour pressure $\left(\frac{{\mathrm{p}}^0-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^0}\right)$
ii) According to Raoult's law the relative lowering of vapour PRESSURE of a dilute solution containing non-volatile solute is equal to the mole fraction of the solute
iii) The molecular weight of solute can be calculated as follows.
$\text{ i.e }\frac{{\mathrm{p}}^0-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^0}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}\left(\text{ since }{\mathrm{x}}_2=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1+{\mathrm{n}}_2}\right)$
Where n₁ and n₂ are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n₂ < < n₁, hence n₂ can be neglected in the denominator
$\begin{array}{l}\frac{{\mathrm{p}}^0-{\mathrm{p}}_{\mathrm{s}}}{{\mathrm{p}}^0}=\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\\ \frac{{\mathrm{P}}^0-{\mathrm{P}}_{\mathrm{S}}}{{\mathrm{P}}^0}=\frac{{\mathrm{\omega }}_2}{{\mathrm{M}}_2}\times \frac{{\mathrm{M}}_1}{{\mathrm{\omega }}_1}\\ {\mathrm{M}}_2=\frac{{\mathrm{w}}_2\times {\mathrm{M}}_1\times {\mathrm{p}}^0}{{\mathrm{w}}_1\left({\mathrm{p}}^0-{\mathrm{p}}_{\mathrm{s}}\right)}\end{array}$
where
w₁ = wt of solvent
w₂ = wt of solute
M₁ = M.wt of solvent
p₀ = V.P of pure solvent
p_s = V.P of solution
M₂ = molecular weight of unknown solute