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 ($R = 0.083 \, L\, bar\, mol^{-1} \, K^{-1}$):

• A. $61038 \,g\, mol^{-1}$
• B. \(51022 \,g\, mol^{-1}\)
• C. $122044 \,g \,mol^{-1}$
• D. \(31011 \,g \,mol^{-1}\)

Answer 1

The Correct Option is A

To calculate the molar mass of the protein, we'll use the formula for osmotic pressure: 

\[\Pi = \dfrac{n}{V}RT\]

,

where:

• \(\Pi\) is the osmotic pressure.
• \(n\) is the number of moles of solute.
• \(V\) is the volume of solution in liters.
• \(R\) is the universal gas constant (0.083 L bar mol^-1 K^-1).
• \(T\) is the temperature in Kelvin.

First, convert the volume from mL to L: \(200 \, \text{mL} = 0.2 \, \text{L}\).

The number of moles, \(n\), is given by: 

\[n = \dfrac{\text{mass}}{\text{Molar Mass}} = \dfrac{1.26\, \text{g}}{M}\]

.

Substitute these values into the osmotic pressure equation: 

\[\Pi = \dfrac{1.26}{M \times 0.2} \times 0.083 \times 300\]

Given that the osmotic pressure \(\Pi\) is \(2.57 \times 10^{-3}\) bar, rearrange the formula to solve for \(M\):

\[2.57 \times 10^{-3} = \left(\dfrac{1.26}{M \times 0.2}\right) \times 0.083 \times 300\]\]

Simplifying and solving for \(M\):

\[2.57 \times 10^{-3} = \left(\dfrac{1.26 \times 0.083 \times 300}{M \times 0.2}\right)\]\]

\[ \therefore M = \dfrac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.2} \]

Calculate the value of \(M\):

\[M = \dfrac{1.26 \times 24.9}{0.000514} \]\]

\[ M ≈ \dfrac{31.374}{0.000514} \]

\[ M ≈ 61038 \, \text{g/mol} \]

Hence, the molar mass of the protein is approximately \(61038 \, \text{g/mol}\), which matches the option 61038 g/mol.