At 300K, the osmotic pressue of 300mL of a protein aqueous solution is $8.3\times {10}^{-5}$ bar. The molar mass of protein is ${10}^4{\mathrm{gmol}}^{-1}$.The weight (in g) of the protein present in this solution is nearly ?

$\left(\mathrm{R}=0.083 \mathrm{L}\text{ bar }{\mathrm{mol}}^{-1}{ \mathrm{K}}^{-1}\right)$

We know osmotic pressure can be mathematically represented as:

$Osmotic Pressure=\pi =\mathrm{CRT}$

Substituting the given values in the question to the formula we obtain:

$$\begin{gathered} \Rightarrow 8.3\times {10}^{-5}=\mathrm{C}\times 0.083\times 300 \\ \mathrm{C}=\frac{{10}^{-3}}{300}=\frac{1}{3}\times {10}^{-5} \\ \Rightarrow \frac{\text{ moles }}{\text{ vol }}=\frac{1}{3}\times {10}^{-5} \\ \Rightarrow \frac{\mathrm{n}}{0.3}=\frac{{10}^{-5}}{3} \\ \Rightarrow \mathrm{n}={10}^{-6} \\ \Rightarrow \frac{\text{ mass }}{\text{ molar mass }}={10}^{-6} \\ \Rightarrow \text{ mass }={10}^{-6}\times {10}^4 \\ =0.01 \end{gathered}$$