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200 cm3 of an aqueous 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 × 10-3 bar. Calculate the molar mass of the protein.

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\(\pi = \frac{W_BRT}{M_BV}\)

\(M_B= \frac{W_BRT}{\pi V}\)

\(\frac{1.26\,g\,\times\,0.083\,L\,bar\,K^{-1}\,mol^{-1}\times300\,K}{2.57\times10^{-3}\,bar\times0.2\,L}\)

= 61,039 g mol-1

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