18g of glucose is dissolved in 90g of water .Then the relative lowering of vapour pressure of
solution is

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

$\frac{∆\mathrm{p}}{{\mathrm{P}}^0}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}$

Wt of Glucose 1 = W =18g
2 Wt of water = W = 90g
2 GMW of water = M =18g

No. of moles of Glucose ${\mathrm{n}}_1=\frac{{\mathrm{W}}_1}{{\mathrm{M}}_1}\frac{18}{180}=\mathrm{O}.\mathrm{I}$

No. of moles of Glucose ${\mathrm{n}}_1=\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2}=\frac{90}{18}=5$

$\mathrm{R}.\mathrm{L}.\mathrm{V}.\mathrm{P}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_1+{\mathrm{n}}_2}=\frac{0.1}{0.1+5}=\frac{0.1}{5.1}=\frac{1}{51}$