Question

Question 7 of an unknown protein are dissolved in enough solvent to make of solution. The osmotic pressure of this solution is measured to be at . Calculate the molar mass of the protein. Round your answer to significant digits.

Answer 1

The question is incomplete, here is the complete question:

371. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.118 atm at 25°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.

Answer: The molar mass of the unknown protein is $1.54\times 10^3g/mol$

Explanation:

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

$\pi=iMRT$

or,

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

where,

$\pi$ = osmotic pressure of the solution = 0.118 atm

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

Mass of protein = 371. mg = 0.371 g   (Conversion factor:  1 g = 1000 mg)

Molar mass of protein = ?

Volume of solution = 5.00 mL

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

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

Putting values in above equation, we get:

$0.118atm=1\times \frac{0.371\times 1000}{\text{Molar mass of protein}\times 5}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=\frac{1\times 0.371\times 1000\times 0.0821\times 298}{0.118\times 5}=1538.4g/mol=1.54\times 10^3g/mol$

Hence, the molar mass of the unknown protein is $1.54\times 10^3g/mol$