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 × 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1):

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 × 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1): (A) 61038 g mol-1 (B) 51022 g mol -1 (C) 122044 g mol-1 (D) 31011 g mol -1

Tardigrade Admin · Accepted Answer

π v = (w/m)RT 2.57 ×10-3 ×(200/1000) = (1.26/m) ×0.083 ×300 m = 61038 gm mol-1

Q. $200 m L$ 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 $2.57 \times 1 0^{- 3}$ bar.
The molar mass of protein will be ( $R = 0.083 L ba r m o l^{- 1} K^{- 1}$ ):

$61038 g m o l^{- 1}$

$51022 g m o l^{- 1}$

$122044 g m o l^{- 1}$

$31011 g m o l^{- 1}$

$π v = \frac{w}{m} RT$
$2.57 \times 1 0^{- 3} \times \frac{200}{1000} = \frac{1.26}{m} \times 0.083 \times 300$
$m = 61038 g m m o l^{- 1}$

Q. 200mL of an aqueous solution of a protein contains its 1.26g. The Osmotic pressure of this solution at 300K is found to be 2.57×10−3 bar. The molar mass of protein will be (R=0.083Lbarmol−1K−1):

61038gmol−1

51022gmol−1

122044gmol−1

31011gmol−1