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Osmotic Pressure

0.2 L of aque...

Question

0.2 L of 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 $2.57 \times 10^{- 3}$ bar. Calculate the molar mass of the protein. $(R = 0.083 L b a r m o l^{- 1} K^{- 1})$ .

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Solution

The expression for the molar mass of the protein is:
$M_{2} = \frac{W_{2}}{π} \frac{R T}{V}$
$M_{2} = \frac{1.26 g}{2.57 \times 10^{- 3} b a r .} \frac{0.0821 L b a r / m o l / K \times 300 K}{0.2 L}$
$M_{2} = 61038 g / m o l$
Thus, the molar mass of the protein is 61038 g/mol.

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