Chemistry - Liquid Solution Question with Solution | TestHub

The osmotic pressure of a non-electrolytic solution can be calculated as follows,

$\mathrm{\pi }=\mathrm{CRT}$

$\mathrm{\pi }$= osmotic pressure

$\mathrm{C}$ = molarity

$\frac{400 \mathrm{Pa}}{{10}^5}=\frac{\frac{2.5 \mathrm{g}}{{\mathrm{M}}_{\mathrm{o}}}}{250/1000 \mathrm{L}}\times 0.83\frac{\mathrm{L}-\mathrm{bar}}{\mathrm{K}.\mathrm{mol}}\times 300 \mathrm{K}$

$\Rightarrow \frac{400 \times {10}^{-5}}{(1.01325)} = \left[\frac{2.5 \times 1000}{{\mathrm{M}}_{\mathrm{o}} \times 250}\right] 0.083 \times 300$

$\Rightarrow 400 \times {10}^{-5} = \frac{10}{{\mathrm{M}}_{\mathrm{o}}} \times 0.083 \times 300$

$\Rightarrow {\mathrm{M}}_{\mathrm{o}} = \frac{10 \times 0.83 \times 300}{400} \times {10}^5$

$\Rightarrow {\mathrm{M}}_{\mathrm{o}} = \left[\frac{10 \times 0.083 \times 3}{4}\right] \times {10}^5$

$\Rightarrow {\mathrm{M}}_{\mathrm{o}}=62250$ gram/mole