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Calculate osmotic pressure of solution of 0.025 mole glucose in $100 \mathrm{~mL}$ water at $300 \mathrm{~K}$. $\left[\mathrm{R}=0.082 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]$
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The correct answer is:
$6.15 \mathrm{~atm}$
$\begin{aligned} \pi & =\mathrm{MRT}=\frac{\mathrm{n}_2 \mathrm{RT}}{\mathrm{V}} \\ & =\frac{0.025 \mathrm{~mol} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}}{0.1 \mathrm{dm}^3} \\ & =6.15 \mathrm{~atm}\end{aligned}$
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