Chemistry: 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.

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371. mg of an unknown protein are dissolved in enough

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myaj580

16.01.2022

Chemistry

The molar mass of the protein is 15394.2 g/mol

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:

$0.118atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{0.118}{1\times 0.0821\times 298}=4.82\times 10^{-3}M$

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

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

We are given:

Molarity of solution = $4.82\times 10^{-3}M$

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:

$4.82\times 10^{-3}M=\frac{0.371\times 1000}{\text{Molar mass of protein}\times 5.00}\\\\\text{Molar mass of protein}=\frac{0.371\times 1000}{4.82\times 10^{-3}\times 5.00}=15394.2g/mol$

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

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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.

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