Question

Find the osmotic pressure of $0.05\text{M}$ solution of urea at ${27}^{\circ }C$
$\text{Molecular formula of urea is}\left(C{H}_4{N}_{2}O\right)$

• A. $1.2\text{atm}$
• B. $2.5\text{atm}$
• C. $0.1\text{atm}$
• D. $0.7\text{atm}$

Answer 1

The correct option is A $1.2\text{atm}$

Osmotic pressure: $\left(\pi \right)$
$\pi =C\times R\times T$
Here
$C\text{is concentration (Molarity)}{\text{(mol L}}^{-1}\left)R\text{is ideal gas constant}{\text{(L atm mol}}^{-1}{\text{K}}^{-1}\right)T\text{is temperature (K)}$
Given,
$C=0.05{\text{mol L}}^{-1}R=0.082{\text{L atm mol}}^{-1}{\text{K}}^{-1}T={27}^{\circ }C=300\text{K}$

$\therefore \pi =(0.05\times 0.082\times 300)atm\Rightarrow \pi =1.23\text{atm}$