Calculate osmotic pressure of solution of 0.025 mole glucose in $100 \mathrm{~mL}$ water at $300…

Calculate osmotic pressure of solution of 0.025 mole glucose in $100 \mathrm{~mL}$ water at $300 \mathrm{~K}$. $\left[\mathrm{R}=0.082 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]$
  1. $1.54 \mathrm{~atm}$
  2. $2.05 \mathrm{~atm}$
  3. $6.15 \mathrm{~atm}$
  4. $3.08 \mathrm{~atm}$

Solution

$\begin{aligned} \pi & =\mathrm{MRT}=\frac{\mathrm{n}_2 \mathrm{RT}}{\mathrm{V}} \\ & =\frac{0.025 \mathrm{~mol} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}}{0.1 \mathrm{dm}^3} \\ & =6.15 \mathrm{~atm}\end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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