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 at 0.118 atm at 25 C

Calculate the molar mass of the protein.

Answer 1

mm protein = 15365.8183 g/mol

Step-by-step explanation:

∴ molar mass (mm) ≡ g/mol

• osmotic pressure (π) = C RT

∴ π = 0.118 atm

∴ T = 25°C ≅ 298 K

∴ concentration (C ) [=] mol/L

∴ mass protein = 371 mg = 0.371 g

∴ volume sln = 5.00 mL = 5 E-3 L

⇒ C = π / RT = (0.118 atm)/((0.082 atm.L/K.mol)(298 K))

⇒ C = 4.8289 E-3 mol/L

⇒ mol protein = (4.8289 E-3 mol/L)×(5 E-3 L) = 2.4145 E-5 mol

⇒ mm protein = (0.371 g)/(2.4145 E-5 mol) = 15365.8183 g/mol

Answer 2

Answer: The molar mass of the protein is 15394.2 g/mol

Step-by-step explanation:

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

`\pi=iMRT`

where,

`\pi` = osmotic pressure of the solution = 0.118 atm

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

M = molarity of solute = ?

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:

`0.118atm=1* M* 0.0821\text{ L.atm }mol^(-1)K^(-1)* 298K\\\\M=(0.118)/(1* 0.0821* 298)=4.82* 10^(-3)M`

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

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

We are given:

Molarity of solution =
`4.82* 10^(-3)M`

Given mass of protein = 371. mg = 0.371 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 5.00 mL

Putting values in above equation, we get:

`4.82* 10^(-3)M=\frac{0.371* 1000}{\text{Molar mass of protein}* 5.00}\\\\\text{Molar mass of protein}=(0.371* 1000)/(4.82* 10^(-3)* 5.00)=15394.2g/mol`

Hence, the molar mass of the protein is 15394.2 g/mol