Question

428. mg 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:

428. 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 0.0766 atm at 25.0°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits

Answer: The molar mass of protein is
`27.3* 10^3g/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 solute (protein) = 428 mg = 0.428 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:

`0.0766=1* \frac{0.428* 1000}{\text{Molar mass of protein}* 5}* 0.0821\text{ L. atm }mol^(-1)K^(-1)* 298K\\\\\text{Molar mass of protein}=(1* 0.428* 1000* 0.0821* 298)/(0.0766* 5)=27340.4g/mol=27.3* 10^3g/mol`

Hence, the molar mass of protein is
`27.3* 10^3g/mol`