6g of non volatile solute is dissolved in 90g of water, such that the lowering of vapour pressure is 2% . The molecular weight of solute is

Wt of non volatile solute ${\mathrm{W}}_1$  = 6g

Wt of solvent (Water)=${\mathrm{W}}_2$=90g

GMW of solute$={\mathrm{M}}_1=?$

GMW of solvent $\left({\mathrm{H}}_2\mathrm{O}\right)=M_2=18g$

$\mathrm{R}.\mathrm{L}.\mathrm{V}.\mathrm{P}=\frac{∆\mathrm{p}}{{\mathrm{p}}^0}=\frac{2}{100}$

$\mathrm{According} \mathrm{to} \mathrm{Raoult}’\mathrm{s} \mathrm{law}$

$\frac{∆\mathrm{p}}{{\mathrm{p}}^0}=\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1}\times \frac{{\mathrm{M}}_2}{{\mathrm{W}}_2}$

$\frac{2}{100}=\frac{6}{M_1}\times \frac{18}{90}$

${M}_1=\frac{6\times 18}{2\times 90}\times 100; M_1=60$