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A solution of $8 \mathrm{~g}$ of certain organic compound in $2 \mathrm{dm}^3$ water develops osmotic pressure $0.6 \mathrm{~atm}$ at $300 \mathrm{~K}$. Calculate the molar mass of compound. $\left[\mathrm{R}=0.082 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
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$164 \mathrm{~g} \mathrm{~mol}^{-1}$
$\begin{aligned} \pi & =\frac{\mathrm{W}_2 \mathrm{RT}}{\mathrm{M}_2 \mathrm{~V}} \\ \therefore \quad \mathrm{M}_2 & =\frac{\mathrm{W}_2 \mathrm{RT}}{\pi \mathrm{V}} \\ \therefore \quad \mathrm{M}_2 & =\frac{8 \mathrm{~g} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.6 \mathrm{~atm} \times 2 \mathrm{dm}^3} \\ & =164 \mathrm{~g} \mathrm{~mol}^{-1}\end{aligned}$
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