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. $31011gmo{l}^{-1}$
• B. $51022gmo{l}^{-1}$
• C. $122044gmo{l}^{-1}$
• D. $61038gmo{l}^{-1}$

Answer 1

The correct option is D $61038gmo{l}^{-1}$

$\pi V=\frac{w}{m}RT$
$\pi \text{is osmotic pressure}$
$T\text{is temperature}$
$R\text{is ideal gas constant}$
$w\text{is mass}$
$m\text{is molar mass}$
$V\text{is volume in L}$

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

$=61038gmo{l}^{-1}$

Hence, option (a) is correct.