1.78 g of an optically active L-amino acid (A) is treated with NaNO2/HCl at 0 ° C. 448 cm3 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

Question

1.78 g of an optically active L-amino acid (A) is treated with NaNO2/HCl at 0 ° C. 448 cm3 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 (A) 34,500 g mol-1 (B) 34,600 g mol-1 (C) 36,500 g mol -1 (D) 35,600 g mol -1

Tardigrade Admin · Accepted Answer

From van-Siyke method for estimation of amino acids

Molar mass of L-amino acid

= \frac{1.78 \times 22400}{448}

= 89

∵

Protein has

0.25%

of amino acid by mass

∴

Molar mass of protein

= \frac{89 \times 100}{0.25}

= 35600 g m o l^{- 1}

Q. $1.78 g$ of an optically active L-amino acid $(A)$ is treated with $N a N O_{2} / H Cl$ at $0^{\circ} C$ . $448 c m^{3}$ 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

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

$34, 600 g m o l^{- 1}$

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

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

From van-Siyke method for estimation of amino acids

Molar mass of L-amino acid $= \frac{1.78 \times 22400}{448}$

$= 89$

$∵$ Protein has $0.25%$ of amino acid by mass

$∴$ Molar mass of protein $= \frac{89 \times 100}{0.25}$

$= 35600 g m o l^{- 1}$

Q. 1.78g of an optically active L-amino acid (A) is treated with NaNO2​/HCl at 0∘C. 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

34,500gmol−1

34,600gmol−1

36,500gmol−1

35,600gmol−1