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Question

200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57×103 bar. Calculate the molar mass of the protein.

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Solution

Given:
Osmotic pressure (π)=2.57×103 bar,
Volume of solution (V)=200 cm3=0.2 litre,
Temperature (T)=300 K
And mass of solute (protein) =1.26 g

We know that universal gas constant (R)=0.083 L bar mol1K1

We know that osmotic pressure (π)=CRT
And we also know that molarity (C)=mass of solute(m_2)molar mass of solute(M_2)×volume of solution(L)
So, π=mass of solute(m_2)×R×Tmolar mass of solute(M_2)×volume of solution(L)

Molar mass of solute (M2)=1.26g×0.083 L K1 mol1×300 K2.57×103 bar×0.2L

=61,038.91gmol1

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