Question

200 mL of an aqueous solution of a protein contains its 1.26 g . The osmotic pressure of this solution at 300 K is found to be 2.57 x 10^-3 bar. The molar mass of protein will be (R = 0.083 L bar mol^-1 K^-1)

51022 g mol^-1

122044 g mol^-1

31011 g mol^-1

61038 g mol^-1

Solution

61038 g mol^-1

straight pi space equals space CRT equals space fraction numerator straight w space straight x space 1000 over denominator straight M space straight x space straight V space left parenthesis in space mL right parenthesis end fraction space straight x space RT open square brackets therefore space space straight C space equals space fraction numerator straight n space straight x space 1000 over denominator straight V space in space mL end fraction and space straight n space equals space straight w over straight M close square brackets equals space fraction numerator 1.26 space straight x space 1000 space straight x space 0.083 space straight x space 300 over denominator 2.57 space straight x space 10 to the power of negative 3 end exponent space straight x space 200 end fraction equals space 61038 space straight g space mol to the power of negative 1 end exponent

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