Question

A protein subunit from an enzyme is part of a research study and needs to be characterized. A total of 0.150 g of this subunit was dissolved in enough water to produce 2.00 mL of solution. At 28 ∘C the osmotic pressure produced by the solution was 0.138 atm. What is the molar mass of the protein?

Answer 1

Answer: The molar mass of the protein is 13392.86 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 0.138 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $28^oC=[273+28]=301K$

Putting values in above equation, we get:

$0.138atm=1\times M\times 0.0820\text{ L.atm }mol^{-1}K^{-1}\times 301K\\\\c=0.0056M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.0056 M

Given mass of protein = 0.150 g

Volume of solution = 2 mL

Putting values in above equation, we get:

$0.0056M=\frac{0.150\times 1000}{\text{Molar mass of protein}\times 2}\\\\\text{Molar mass of protein}=13392.86g/mol$

Hence, the molar mass of the protein is 13392.86 g/mol