Answer: The molar mass of the insulin is 6087.2 g/mol
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 = 15.5 mmHg
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (insulin) = 33 mg = 0.033 g (Conversion factor: 1 g = 1000 mg)
Volume of solution = 6.5 mL
R = Gas constant = [tex]62.364\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]25^oC=[273+25]=298K[/tex]
Putting values in above equation, we get:
[tex]15.5mmHg=1\times \frac{0.033\times 1000}{\text{Molar mass of insulin}\times 6.5}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of insulin}=\frac{1\times 0.033\times 1000\times 62.364\times 298}{15.5\times 6.5}=6087.2g/mol[/tex]
Hence, the molar mass of the insulin is 6087.2 g/mol
