Question

1.0 g of a protein is dissolved in 100 ml of distilled water. The resulting solution has an osmotic pressure of 0.001 atm at room temperature. What is the molecular weight of the protein?

Answer 1

Answer: The molar mass of the protein is 244678.2 g/mol

Explanation:

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

$\pi=icRT$

where,

$\pi$ = osmotic pressure of the solution = 0.001 atm

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

c = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$0.001atm=1\times c\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\c=\frac{0.001}{1\times 0.0821\times 298}=4.087\times 10^{-5}M$

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 = $4.087\times 10^{-5}M$

Given mass of protein = 1.0 grams

Volume of solution = 5.00 mL  

Putting values in above equation, we get:  

$4.087\times 10^{-5}M=\frac{1.0\times 1000}{\text{Molar mass of protein}\times 100}\\\\\text{Molar mass of protein}=\frac{1\times 1000}{4.087\times 10^{-5}\times 100}=244678.2g/mol$

Hence, the molar mass of the protein is 244678.2 g/mol