Question

of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measured to be 0.107atm at 25.0°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.

Answer 1

The question is incomplete . The complete question is :

100 mg of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measured to be 0.107atm at 25.0°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.

Answer:  The molar mass of the protein is $4.57\times 10^3g/mol$

Explanation:

$\pi =CRT$

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

where,

$\pi$ = osmotic pressure of the solution = 0.107 atm

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

Mass of solute (protein) = 100 mg = 0.1 g   (Conversion factor: 1 g = 1000 mg)

Volume of solution = 5.00 mL

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

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

Putting values in above equation, we get:

$0.107=1\times \frac{0.1\times 1000}{\text{Molar mass of insulin}\times 5.00}\times 0.0821\text{ Latm }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of protein}$

$\text{molar mass of protein}=4.57\times 10^3g/mol$

Hence, the molar mass of the protein is $4.57\times 10^3g/mol$