# 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

1. A

180

2. B

90

3. C

270

4. D

200

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### Solution:

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

$\frac{{\mathrm{P}}^{0}-\mathrm{P}}{{\mathrm{P}}^{0}}=\frac{\frac{{W}_{1}}{{M}_{1}}}{\frac{{W}_{1}}{{M}_{1}}+\frac{{W}_{2}}{{M}_{2}}}$

$\frac{3000-2985}{3000}=\frac{\frac{5}{{M}_{1}}}{\frac{5}{{M}_{1}}+\frac{100}{18}}$

$\frac{15}{3000}=\frac{\frac{5}{{M}_{1}}}{\frac{90+100{M}_{1}}{18{M}_{1}}}$

$\frac{15}{3000}=\frac{5}{\overline{){\mathrm{M}}_{1}}}×\frac{18\overline{){\mathrm{M}}_{1}}}{\left(90+100{\mathrm{M}}_{1\right)}}$

$90+100{\mathrm{M}}_{1}=\frac{5×18×3000}{15}$

$100{\mathrm{M}}_{1}=\frac{15000×18}{15}-90$

$100{\mathrm{M}}_{1}=18000-90$

${\mathrm{M}}_{1}=\frac{17910}{100}$

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