Question

200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be $2.57\times {10}^{-3}$ bar. The molar mass of protein will be:

($R=0.083Lbarmo{l}^{-1}{K}^{-1}$)

• A. 61038 g mol${}^{-1}$
• B. 51022 g mol${}^{-1}$
• C. 122044 g mol${}^{-1}$
• D. 31011 g mol${}^{-1}$

Answer 1

Answer: (A)

61038 g mol${}^{-1}$

$\pi v=\frac{w}{m}RT$

$2.57\times {10}^{-3}\times \frac{200}{1000}=\frac{1.26}{m}\times 0.083\times 300$

$m=61038$ gm mol${}^{-1}$

Hence, the correct option is $A$