Northern cod produce proteins that protect their cells from damage caused by subzero temperatures. Measurements of the osmotic pressure for two "antifreeze" proteins at 18Â°C yielded the data listed below. Use this information to calculate the molar mass for each of the proteins. Assume these proteins are nonelectrolytes and use the value i = 1.

Required:
If a 54.1 mg sample of protein A in 1.5 mL of water has an osmotic pressure of 0.285 atm, what is the molar mass of protein A?

[tex]\pi=i\times \frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution}}\times R\times T[/tex] .....(1)

Where,

[tex]\pi[/tex] = osmotic pressure = 0.285 atm

i = 1

Given mass of protien (solute) = 54.1 mg = 0.0541 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 1.5 mL

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature = [tex]18^oC=[18+273]=291K[/tex]

Plugging values in equation 1:

[tex]0.285atm=1\times \frac{0.0541\times 1000}{\text{Molar mass of protein}\times 1.5}\times 0.0821L.atm/mol.K\times 291K\\\\\text{Molar mass of protein}=\frac{1\times 0.0541\times 1000\times 0.0821\times 291}{0.285\times 1.5}\\\\\text{Molar mass of protein}=3023.41g/mol[/tex]

Hence, the molar mass of protein A is 3023.41 g/mol.

pstnonsonjoku pstnonsonjoku · Answer 2 · 2022-02-08T15:44:22+01:00

The molar mass of the protein is 3018 g/mol.

We know that the osmotic pressure is obtained using the formula;

Osmotic pressure = iMRT

i = Van't Hoff factor
M = molar concentration
R = gas constant
T = absolute temperature

concentration of the protein = 54.1 * 10^-3g/M/ 1.5 * 10^-3 L

Where M = molar mass of the protein

So;

0.285 atm = 54.1 * 10^-3g/ 1.5 * 10^-3 L M * 0.0821 L.atm/mol.K * 291 K

0.285 = 1.29/1.5 * 10^-3M

M = 1.29/0.285 * 1.5 * 10^-3

M = 3018 g/mol

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