The vapour pressure of a solution of 5g of non electrolyte in 100g of water at a particular temperature is 2 2985 ${\mathrm{Nm}}^{-2}$ . The vapour pressure of pure water at that temperature is 3000 ${\mathrm{Nm}}^{-2}$.The molecular weight of the solute is

Vapour pressure of solution=P=2985${\mathrm{Nm}}^{-2}$

Vapour pressure of water$={\mathrm{P}}^0=300{\mathrm{Nm}}^{-2}$

Wt. of non electrolyte solute$={\mathrm{W}}_1=5\mathrm{g}$

Wt. of water$={\mathrm{W}}_2=100\mathrm{g}$

GMW of water$={\mathrm{M}}_2=18\mathrm{g}$

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

According to Raoult’s law

$\frac{{\mathrm{P}}^0-\mathrm{P}}{{\mathrm{P}}^0}=\frac{\frac{W_1}{M_1}}{\frac{W_1}{M_1}+\frac{W_2}{M_2}}$

$\frac{3000-2985}{3000}=\frac{{\displaystyle \frac{5}{M_1}}}{{\displaystyle \frac{5}{M_1}}+{\displaystyle \frac{100}{18}}}$

$\frac{15}{3000}=\frac{{\displaystyle \frac{5}{M_1}}}{{\displaystyle \frac{90+100M_1}{18M_1}}}$

$\frac{15}{3000}=\frac{5}{\cancel{{\mathrm{M}}_1}}\times \frac{18\cancel{{\mathrm{M}}_1}}{(90+100{\mathrm{M}}_{1)}}$

$90+100{\mathrm{M}}_1=\frac{5\times 18\times 3000}{15}$

$100{\mathrm{M}}_1=\frac{15000\times 18}{15}-90$

$100{\mathrm{M}}_1=18000-90$

${\mathrm{M}}_1=\frac{17910}{100}$

${\mathrm{M}}_1=179.1(\mathrm{Near} \mathrm{By})$