Respuesta :
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 [tex]1.54\times 10^3g/mol[/tex]
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = 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 = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]25^oC=[25+273]K=298K[/tex]
Putting values in above equation, we get:
[tex]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[/tex]
Hence, the molar mass of the unknown protein is [tex]1.54\times 10^3g/mol[/tex]
