Question

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 0.0766 atm at 25.0 degree C .

Calculate the molar mass of the protein. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: The molar mass of the unknown protein is 6387.9 g/mol

Step-by-step explanation:

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

`\pi=iMRT`

or,

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

where,

`\pi` = osmotic pressure of the solution = 0.0766 atm

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

Mass of protein = 100. mg = 0.100 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:

`0.0766atm=1* \frac{0.100* 1000}{\text{Molar mass of protein}* 5}* 0.0821\text{ L. atm }mol^(-1)K^(-1)* 298K\\\\\pi=(1* 0.100* 1000* 0.0821* 298)/(0.0766* 5)=6387.9g/mol`

Hence, the molar mass of the unknown protein is 6387.9 g/mol