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Calculate the amount of solute dissolved in $3 \mathrm{dm}^3$ water having osmotic pressure $0.3 \mathrm{~atm}$ at $300 \mathrm{~K}$
(Molar mass of solute $=108 \mathrm{~g} \mathrm{~mol}^{-1}, R=0.0821 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )
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(Molar mass of solute $=108 \mathrm{~g} \mathrm{~mol}^{-1}, R=0.0821 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )
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The correct answer is:
3.95 gram
Osmotic pressure
$\begin{aligned} & \pi=i C R T \\ & 0.3=1 \times \frac{W}{108} \times \frac{1}{3} \times 0.0821 \times 300 \\ & W=3.95 \mathrm{gm}\end{aligned}$
$\begin{aligned} & \pi=i C R T \\ & 0.3=1 \times \frac{W}{108} \times \frac{1}{3} \times 0.0821 \times 300 \\ & W=3.95 \mathrm{gm}\end{aligned}$
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