Calculate osmotic pressure exerted by a solution containing $0.822 \mathrm{~g}$ of solute in $300…

Calculate osmotic pressure exerted by a solution containing $0.822 \mathrm{~g}$ of solute in $300 \mathrm{~mL}$ of water at $300 \mathrm{~K}$. (Molar mass of solute $=340 \mathrm{~mol}^{-1}, \mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )
  1. 0.5 atm
  2. 0.2 atm
  3. 0.1 atm
  4. 0.4 atm

Solution

$\begin{aligned} & \pi=\mathrm{CRT} \\ & \mathrm{C}=\frac{\text { moles of solute }}{\text { volumeof solution }(\mathrm{ml})} \times 1000 \\ & =\frac{0.822 / 340}{300} \times 1000 \\ & =0.008 \\ & \pi=0.008 \times 0.0821 \times 300 \\ & \pi=0.2 \mathrm{~atm}\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

Practice more Solutions questions on Aicharya