2.5 g of protein containing only glycine (C₂H₅NO₂) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be $5 .03 \times 10^{- 3} bar$ . The total number of glycine units present in the protein is __________.
(Given : R = 0.083 L bar K^-1 mol^-1)

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answer is 330.

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Detailed Solution

$\begin{array}{l} π = CRT \\ 5 .03 \times 10^{- 3} = C \times 0 .083 \times 300 \\ \begin{array}{l} C = 0 .202 \times 10^{- 3} M \\ Moles of protein = 0 .202 \times 10^{- 3} \times 0 .5 \\ = 10^{- 4} \times 1 .01 \\ 1 .01 \times 10^{- 4} = \frac{2 .5}{M} \end{array} \\ M(molar mass of protein) = 24752 \\ ∴ No. of glycine units = \frac{24752}{75} = 330 .03 \end{array}$