At $300 \mathrm{~K}, 6 \mathrm{~g}$ of urea was dissolved in 500 mL of water. What is the osmotic pressure (in atm) of resultant solution? $\begin{aligned} & \left(R=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ight) \ & (\mathrm{C}=12 ; \mathrm{N}=14 ; O=16 ;…

Question

At $300 \mathrm{~K}, 6 \mathrm{~g}$ of urea was dissolved in 500 mL of water. What is the osmotic pressure (in atm) of resultant solution? $\begin{aligned} & \left(R=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}ight) \ & (\mathrm{C}=12 ; \mathrm{N}=14 ; O=16 ; \mathrm{H}=1) \end{aligned}$

Aicharya · Accepted Answer

Osmotic pressure $(\pi)=$ CRT Given,
Weight of urea $=6 \mathrm{~g}$ Molecular weight of urea

$\therefore \quad$ Number of mole of urea $=\frac{\text { Weight }}{\text { Molecular weight }}$ $\Rightarrow \frac{6}{60}=0.1 \mathrm{~mole}$ $\begin{aligned} & \operatorname{concentration}(\mathrm{C})=\frac{\text { mole }}{\text { volume }} \\ & \\ & \qquad=\frac{0.1}{0.5} \quad\left[1 \mathrm{~L}=10^3 \mathrm{~mL}\right] \end{aligned}$ (C) $=0.2 \mathrm{~mole} / \mathrm{liter}$. $\begin{aligned} \pi & =0.2 \times 0.082 \times 300 \\ \pi & =4.92 \mathrm{~atm} \end{aligned}$

At $300 \mathrm{~K}, 6 \mathrm{~g}$ of urea was dissolved in 500 mL of water. What is the osmotic pressure…

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