Question

At $300$ K, the osmotic pressure of $300$ mL of a protein aqueous solution is $8.3\times {10}^{-5}$ bar. The molar mass of protein is ${10}^4$ g/mol. What is the weight (in g) of the protein present in this solution? (R=0.083 L bar/mol K)

• A. $0.1$
• B. $1.0$
• C. $10$
• D. $0.01$

Answer 1

Answer: (D)

$0.01$

$OP=\pi =CRT$

$\Rightarrow 8.3\times {10}^{-5}=C\times 0.083\times 300$

$C=\frac{{10}^{-3}}{300}=\frac{1}{3}\times {10}^{-5}$

$\Rightarrow \frac{moles}{vol}=\frac{1}{3}\times {10}^{-5}$

$\Rightarrow \frac{n}{0.3}=\frac{{10}^{-5}}{3}$

$\Rightarrow n={10}^{-6}$

$\Rightarrow \frac{mass}{molarmass}={10}^{-6}$

$\Rightarrow$ moss $={10}^{-6}\times {10}^4$

$=0.019$