Question

100 cm\(^3\) of an aqueous solution of a protein contains 1.5 g of the protein. The osmotic pressure of such a solution at 300 K is found to be \(4.5 \times 10^{-3}\) bar. The molar mass of the protein is \([R = 0.083 \text{ L atm mol}^{-1}\text{ K}^{-1}]\)}

• A. \(8.3 \times 10^5 \text{ g mol}^{-1}\)
• B. \(4.15 \times 10^4 \text{ g mol}^{-1}\)
• C. \(8.3 \times 10^3 \text{ g mol}^{-1}\)
• D. \(8.3 \times 10^4 \text{ g mol}^{-1}\)
• E. \(4.15 \times 10^4 \text{ g mol}^{-1}\)

Hint:

For molar mass by osmotic pressure: \[ M=\frac{wRT}{\pi V} \] Convert volume into litres before substitution.

Answer 1

The Correct Option is D

Use osmotic pressure formula: \[ \pi = \frac{w}{MV}RT \] So, \[ M = \frac{wRT}{\pi V} \] Given: \[ w=1.5\text{ g} \] \[ V=100\text{ cm}^3 = 0.1\text{ L} \] \[ T=300\text{ K} \] \[ \pi=4.5\times10^{-3}\text{ bar} \] Taking \(R=0.083\), substitute: \[ M=\frac{1.5\times 0.083\times300}{(4.5\times10^{-3})\times0.1} \] \[ =\frac{37.35}{4.5\times10^{-4}} \] \[ =8.3\times10^4\text{ g mol}^{-1} \]
Hence, the correct answer is: \[ \boxed{(D)\ 8.3\times10^4\text{ g mol}^{-1}} \]