At a certain temperature , the vapour pressure of water is 50 mm .The relative lowering of vapour pressure of a solution containing 36 g of glucose in 900 g of water is

Solution: According to Raoult’s Law
R.L.V.P = mole fraction of solute

$\mathrm{R}.\mathrm{L}.\mathrm{V}.\mathrm{P}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}$

Wt of Glucose (solute) = W = 36g
 GMW of Glucose = M =180g
 Wt. of water = W = 900g
 GMW of water = M =18g

No. of Moles of Glucose$={\mathrm{n}}_1=\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1}=\frac{36}{180}=0.2$

No. of Moles of Glucose$={\mathrm{n}}_2=\frac{{\mathrm{W}}_2}{\mathrm{M}2}=\frac{900}{18}=50$

$\mathrm{R}.\mathrm{L}.\mathrm{V}.\mathrm{P}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}$$=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}$$(\therefore {\mathrm{n}}_1<<{\mathrm{n}}_2)$

$=\frac{0.2}{50}=0.004$