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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 $2.57\times10^{-3}$ bar. The molar mass of protein will be (R = 0.083 L bar $mol^{-1}K^{-1}$)
A
31011 g $mol^{-1}$
B
61038 g $mol^{-1}$
C
51022 g $mol^{-1}$
D
122044 g $mol^{-1}$
Detailed Solution
Osmotic pressure: $\pi = CRT = \frac{w}{M}\times\frac{1000}{V(mL)}\times RT$
Rearranging: $M = \frac{w\times1000\times R\times T}{\pi\times V}$
$M = \frac{1.26\times1000\times0.083\times300}{2.57\times10^{-3}\times200}$
Numerator = $1.26\times1000\times24.9 = 31374$
Denominator = $2.57\times10^{-3}\times200 = 0.514$
$M = \frac{31374}{0.514} = 61038$ g $mol^{-1}$
Rearranging: $M = \frac{w\times1000\times R\times T}{\pi\times V}$
$M = \frac{1.26\times1000\times0.083\times300}{2.57\times10^{-3}\times200}$
Numerator = $1.26\times1000\times24.9 = 31374$
Denominator = $2.57\times10^{-3}\times200 = 0.514$
$M = \frac{31374}{0.514} = 61038$ g $mol^{-1}$