Question

$200\text{mL}$ of an aqueous solution of a protein contains its $1.26\text{g}$. The osmotic pressure of this solution at $300\text{K}$ is found to be $2.52\times {10}^{-3}$ bar. The molar mass of protein will be
$(R=0.08{\text{L bar mol}}^{-1}{\text{K}}^{-1})$

• A. $51022{\text{g mol}}^{-1}$
• B. $122044{\text{g mol}}^{-1}$
• C. $31011{\text{g mol}}^{-1}$
• D. $60000{\text{g mol}}^{-1}$

Answer 1

The correct option is D $60000{\text{g mol}}^{-1}$

We know,
Osmotic pressure, $\left(\pi \right)=CRT$
Where,
$C=\text{Concentration}$
$R=\text{Universal gas constant}$
$T=\text{Temperature}$
Let, Molar mass of the protein be $M$
$\therefore 2.52\times {10}^{-3}=\frac{1.26\times 1000}{M\times 200}\times 0.08\times 300$
$\Rightarrow M=\frac{1.26\times 1000\times 0.08\times 300}{2.52\times {10}^{-3}\times 200}=60000{\text{g mol}}^{-1}$