$300 \mathrm{~mL}$ of an aqueous solution of a protein contains $2.52 \mathrm{~g}$ of the protein. If…

$300 \mathrm{~mL}$ of an aqueous solution of a protein contains $2.52 \mathrm{~g}$ of the protein. If osmotic pressure of such a solution at $300 \mathrm{~K}$ is $5.04 \times 10^{-3}$ bar, the molar mass of the protein in $\mathrm{g} \mathrm{mol}^{-1}$ is
  1. $83.0 \times 10^3$
  2. $20.8 \times 10^3$
  3. $41.5 \times 10^3$
  4. $41.5 \times 10^4$

Solution

$p=i \times M \times R \times T$ $ \begin{aligned} & p=\text { osmotic pressure }=5.04 \times 10^{-3} \text { bar }=5.04 \mathrm{~atm} \\ & i=\text { van't Hoff constant }=1 \text { for undissociating } \\ & \text { molecules like protein } \end{aligned} $ molecules like protein $ \begin{aligned} & M=\text { molarity }=\frac{\text { moles solute }}{\text { liter solution }} \\ & =\frac{\text { weight in } \mathrm{g}}{\text { molecular weight }} \times \frac{1}{\text { liter }}=\frac{2.52}{x} \times \frac{1000}{300}=\frac{25.2}{3 x} \end{aligned} $ $ \begin{aligned} T & =\text { Temperature }=300 \mathrm{~K} \\ R & =\text { Gas constant }=0.08206 \mathrm{~L} \mathrm{~atm} / \mathrm{mol} \mathrm{K} . \end{aligned} $ $ \begin{aligned} 5.04 \times 10^{-3} & =1 \times \frac{25.2}{3 x} \times 0.08206 \times 300 \\ 3 x & =\frac{25.2 \times 0.08206 \times 300}{5.04 \times 10^{-3}} \\ x & =\frac{25.2 \times 0.08206 \times 300}{5.04 \times 3 \times 10^{-3}}=\frac{6.2037 \times 100}{1512 \times 10^{-3}} \\ & =0.410 \times 100 \times 10^3=41.0 \times 10^3 \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

Practice more Solutions questions on Aicharya