The osmotic pressure of a solution of NaCl is 0.10 atm and

Solution:

Osmotic pressure $=\pi=\mathrm{i} \times \mathrm{C} \times \mathrm{RT}$

For $\mathrm{NaCl} \mathrm{i}=2$ so

$\pi_{\mathrm{NaCl}}=\mathrm{i} \times \mathrm{C}_{\mathrm{NaCl}} \times \mathrm{RT} \quad \mathrm{C}_{\mathrm{NaCl}}=$ conc. of $\mathrm{NaCl}$

$0.1=2 \times \mathrm{C}_{\mathrm{NaCl}} \times \mathrm{RT}$

$\mathrm{C}_{\text {Glucose }}=\frac{0.2}{\mathrm{RT}}$ $\eta_{\mathrm{NaCl}}=$ No. of moles $\mathrm{NaCl}$

$\eta_{\mathrm{NaCl}}$ in $1 \mathrm{~L}=\mathrm{C}_{\mathrm{NaCl}} \times \mathrm{V}_{\mathrm{Litre}}$

$=\frac{0.05}{\mathrm{RT}} \quad \eta_{\mathrm{glucose}}=$ No. of moles glucose

$\eta_{\text {glucose }}$ in $2 \mathrm{~L}=\mathrm{C}_{\text {glucose }} \times \mathrm{V}_{\text {Litre }}$

$=\frac{0.4}{\mathrm{RT}}$

$\mathrm{V}_{\text {Total }}=1+2=3 \mathrm{~L}$

so Final conc. $\mathrm{NaCl}=\frac{0.05}{3 \mathrm{RT}}$

$\pi_{\text {Total }}=\pi_{\mathrm{NaCl}}+\pi_{\text {glucose }}$

$=\left[\mathrm{i} \times \mathrm{C}_{\mathrm{NaCl}}+\mathrm{C}_{\text {glucose }}\right] \times \mathrm{RT}$

$=\left(\frac{2 \times 0.05}{3 \mathrm{RT}}+\frac{0.4}{3 \mathrm{RT}}\right) \times \mathrm{RT}$

$=\frac{0.5}{3} \mathrm{~atm}$

$=0.1666 \mathrm{~atm}$

$=166.6 \times 10^{-3} \mathrm{~atm}$

$\Rightarrow 167.00 \times 10^{-3} \mathrm{~atm}$

so $x=167.00$