Question

A protein has been isolated as a salt with the formula Na20P (this notation means that there are 20 Na+ ions associated with a negatively charged protein P20−). The osmotic pressure of a 10.0 mL solution containing 0.207 g of the protein is 0.295 atm at 25.0°C. (a) Calculate the molar mass of the protein from these data. g/mol (b) Calculate the actual molar mass of the protein.

Answer 1

Answer:

a) 36,051.81 g/mol is the molar mass of the protein.

b) The actual molar mass of the protein is 757,087.96 g/mol.

Explanation:

Osmotic pressure of the solution = $\pi = 0.295 atm$

Temperature of the solution , T= 25.0°C=25.0+273 K=298 K

Concentration of solution = C

van't Hoff factor = 20 +1 = 21

$\pi =iCRT$

$C=\frac{\pi }{iRT}=\frac{0.295 atm}{21\times 0.0821 atm L/mol K\times 298 K}$

$C=0.0005742 M$

Moles of protein = n

Volume of solution = V = 10.0 mL = 0.010 L

$C=\frac{n}{V(L)}$

$n=0.0005742 M\times 0.010 L=5.742\times 10^{-6}mol$

Mass of  protein = 0.207 g

Molar mass of protein = M

Moles of protein(n) = $\frac{0.207 g}{M}$

$M=\frac{0.207 g}{5.742\times 10^{-6}mol}=36,051.81 g/mol$

36,051.81 g/mol is the molar mass of the protein.

b)

Actual molar mass of protein = M'

van't hoff factor = 21

$i=\frac{M'}{M}$

$M'=21\times 36,051.81 g/mol=757,087.96 g/mol$

The actual molar mass of the protein is 757,087.96 g/mol.