$1\%$ solution of $\mathrm{KCl}$ is dissociated to the extent of $80 \%$. What would be its osmotic pressure at $27 °\mathrm{C}$ $(\mathrm{R}=0.0821\text{ L atm }{\mathrm{K}}^{-} {\text{ mol }}^{-})$

Wt. of solute ${\mathrm{W}}_2=1\mathrm{g}(\because 1\mathrm{\%}\text{ solution });\mathrm{T}$ $=27+273=300\mathrm{K};\mathrm{R}=0.0821{\mathrm{LatmK}}^{-}{\mathrm{mol}}^{-};\mathrm{\alpha }=80\mathrm{\%}=$$\frac{80}{100}=0.8;\text{ volume }=100\mathrm{mL}(\because 1\mathrm{\%}\text{ solution })=$$100\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}=0.1\mathrm{L}\text{; Mol. wt. of }\mathrm{KCl}={\mathrm{M}}_2=39+$$35.5=74.5 \mathrm{g} {\mathrm{mol}}^{-}$

 we know that, Osmotic pressure

$\mathrm{\pi }=\mathrm{iCRT}=\mathrm{i}\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2\mathrm{V}}\times \mathrm{RT}$

For  $$\begin{gathered} \mathrm{KCl} \rightleftharpoons {\mathrm{K}}^{+}+{\mathrm{Cl}}^{-} \\ 1\mathrm{mol} 1\mathrm{mol} \end{gathered}$$

$\mathrm{n}=1+1=2$

(for one K⁺ and on ${\mathrm{CI}}^{-},\mathrm{n}=1+1=2)$

For dissociation, $\mathrm{\alpha }=\frac{\mathrm{i}-1}{\mathrm{n}-1};0.8=\frac{\mathrm{i}-1}{2-1}=\frac{\mathrm{i}-1}{1}=\mathrm{i}-1$

or     $\mathrm{i}=1+0.8=1.8$

But  $\mathrm{\pi }=\frac{{\mathrm{iW}}_2\mathrm{RT}}{{\mathrm{M}}_2\mathrm{V}}$

$\text{ So, }\mathrm{\pi }=\frac{1.8\times 1\mathrm{g}\times 0.0821\mathrm{L} \mathrm{atm} {\mathrm{K}}^{-}{\mathrm{mol}}^{-}\times 300 \mathrm{K}}{74.5 \mathrm{g} {\mathrm{mol}}^{-}\times 0.1\mathrm{L}}$

$\mathrm{\pi }=5.95 \mathrm{atm}$