Q115 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Chemistry
200 mL of an aqueous solution of a protein contain $1.26$ g of protein, the osmotic pressure of this solution at 300 K is found to be $2.57 \times 10^{-3}$ bar, Calculate the molar mass of protein. ( R = 0.083 bar mol$^{-1}$K$^{-1}$)
- A$5102.2$ g mol$^{-1}$
- B$12204.4$ g mol$^{-1}$
- C$31011$ g mol$^{-1}$
- Correct answerD$61039$ g mol$^{-1}$
Solution
Given:
Volume of solution, $V = 200$ mL $= 0.200$ L
Mass of protein, $W = 1.26$ g
Osmotic pressure, $\pi = 2.57 \times 10^{-3}$ bar
Temperature, $T = 300$ K
Gas constant, $R = 0.083$ L bar K$^{-1}$ mol$^{-1}$
Using the formula for osmotic pressure:
$\pi = \dfrac{W}{M V} R T$
Rearranging for molar mass $M$:
$M = \dfrac{W R T}{\pi V}$
Substituting the values:
$M = \dfrac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.200}$
$M = \dfrac{31.374}{5.14 \times 10^{-4}}$
$M = 61038.9$ g mol$^{-1} \approx 61039$ g mol$^{-1}$
Answer: $61039$ g mol$^{-1}$
Volume of solution, $V = 200$ mL $= 0.200$ L
Mass of protein, $W = 1.26$ g
Osmotic pressure, $\pi = 2.57 \times 10^{-3}$ bar
Temperature, $T = 300$ K
Gas constant, $R = 0.083$ L bar K$^{-1}$ mol$^{-1}$
Using the formula for osmotic pressure:
$\pi = \dfrac{W}{M V} R T$
Rearranging for molar mass $M$:
$M = \dfrac{W R T}{\pi V}$
Substituting the values:
$M = \dfrac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.200}$
$M = \dfrac{31.374}{5.14 \times 10^{-4}}$
$M = 61038.9$ g mol$^{-1} \approx 61039$ g mol$^{-1}$
Answer: $61039$ g mol$^{-1}$
Hence, the correct answer is option: 'D'.