$1.78 g$ of an optically active L-amino acid $(A)$ is treated with $N a N O_{2} / H C l$ at $0$ . $448 {c m}^{3}$ of nitrogen was at STP is evolved. A sample of protein has $0.25$ % of this amino acid by mass. The molar mass of the protein is :

34, 500 g {m o l}^{- 1}

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35, 600 g {m o l}^{- 1}

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36, 500 g {m o l}^{- 1}

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35, 400 g {m o l}^{- 1}

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Solution

The correct option is B $35, 600 g {m o l}^{- 1}$
At STP, 1 mole of nitrogen is equal to 22400 cm $^{3}$ .
Hence, 448 cm $^{3}$ of nitrogen is equal to $\frac{448}{22400} = 0.02$ moles.
Number of moles of nitrogen $=$ number of moles of amino acid $=$ 0.02 moles.

A sample of protein has 0.25% of this amino acid by mass.
1.78 g is the mass of amino acid.
Mass of protein $= 1.78 \times \frac{100}{0.25} = 712$ g
Number of moles of protein $=$ 0.02 moles
Molar mass of protein $= \frac{712}{0.02} = 35600 g / m o l$
Hence, the option (B) is the correct answer.

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1.78g of an optically active L-amino acid (A) is treated with NaNO2/HCl at 0. 448cm3 of nitrogen was at STP is evolved. A sample of protein has 0.25% of this amino acid by mass. The molar mass of the protein is :

$1.78 g$ of an optically active L-amino acid $(A)$ is treated with $N a N O_{2} / H C l$ at $0$ . $448 {c m}^{3}$ of nitrogen was at STP is evolved. A sample of protein has $0.25$ % of this amino acid by mass. The molar mass of the protein is :