The osmotic pressure of a solution of a synthetic polyisobutylene in benzene was determined at $25 °C$. A sample containing $0.2 \mathrm{g}$ of solute $\mathrm{per} 100 {\mathrm{cm}}^3$ of solution developed a rise of $2.4 \mathrm{mm}$ at osmotic equilibrium. If density of solution is $0.88 \mathrm{g} {\mathrm{cm}}^{-3}$, the molecular weight of isobutylene is:

Density, $\mathrm{d}=\frac{0.88 \mathrm{g}}{{\mathrm{cm}}^3}=\frac{0.88 \mathrm{g}}{{\mathrm{cm}}^3}\times \frac{1 \mathrm{kg}}{1000 \mathrm{g}}\times \frac{{10}^6 {\mathrm{cm}}^3}{1 {\mathrm{m}}^3}$

$\begin{array}{l}=0.88\times {10}^3 \mathrm{kg} {\mathrm{m}}^{-3}\\ \mathrm{h}=2.4 \mathrm{mm}=2.4 \mathrm{mm}\times \frac{\mathrm{m}}{1000 \mathrm{mm}}\\ =2.4\times {10}^{-3}\mathrm{m};\\ \mathrm{g}=9.81\mathrm{m} {\mathrm{s}}^{-2}\end{array}$

$\therefore$ Osmotic pressure, 

$\begin{array}{l}\mathrm{\pi }=\mathrm{hdg}=2.4\times {10}^{-3}\mathrm{m} \times 0.88\times {10}^3 \mathrm{kg}\\ {\mathrm{m}}^{-3}\times 9.81\mathrm{m} {\mathrm{s}}^{-2}\\ =20.7 \mathrm{kg} {\mathrm{m}}^{-}{\mathrm{s}}^{-2}\end{array}$

We know that $\mathrm{R}=8.314 \mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-2} {\mathrm{K}}^{-}{\mathrm{mol}}^{-}$

$\begin{array}{l}\mathrm{T}=25+273=293 \mathrm{K};\\ 100 {\mathrm{cm}}^3=100 {\mathrm{cm}}^3\times \frac{{\mathrm{m}}^3}{{10}^6 {\mathrm{cm}}^3}={10}^{-4} {\mathrm{m}}^3\end{array}$

But, $\mathrm{\pi }=\frac{\mathrm{nRT}}{\mathrm{V}}\text{ or }\mathrm{\pi }=\frac{{\mathrm{W}}_2}{{\mathrm{M}}_2\mathrm{V}}\times \mathrm{RT}$

 $\begin{array}{l}{\mathrm{M}}_2=\frac{0.2 \mathrm{g} \times 8.314 \mathrm{kg} {\mathrm{m}}^2 {\mathrm{s}}^{-2 }{\mathrm{K}}^{-} {\mathrm{mol}}^{-} \times 293 \mathrm{K}}{20.7 \mathrm{kg} {\mathrm{m}}^{-}{\mathrm{s}}^{-2} \times {10}^{-4 }{\mathrm{m}}^3}\\ =235362.5 \mathrm{g} {\mathrm{mol}}^{-}\end{array}$