Question:
$1.78\, g$ of an optically active L-amino acid $(A)$ is treated with $NaNO_2/HCl$ at $0\,^\circ\,C$. $448\, cm^3$ 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
$1.78\, g$ of an optically active L-amino acid $(A)$ is treated with $NaNO_2/HCl$ at $0\,^\circ\,C$. $448\, cm^3$ 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
Updated On: Feb 23, 2024
- $34,500 \,g \,mol^{-1}$
- $34,600 \,g \,mol^{-1}$
- $36,500 \,g \,mol ^{-1}$
- $35,600\, g\, mol ^{-1}$
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The Correct Option is D
Solution and Explanation
From van-Siyke method for estimation of amino acids
Molar mass of L-amino acid $=\frac{1.78 \times 22400}{448}$
$=89$
$\because$ Protein has $0.25 \%$ of amino acid by mass
$\therefore$ Molar mass of protein $=\frac{89 \times 100}{0.25}$
$=35600\, g\, mol ^{-1}$
Molar mass of L-amino acid $=\frac{1.78 \times 22400}{448}$
$=89$
$\because$ Protein has $0.25 \%$ of amino acid by mass
$\therefore$ Molar mass of protein $=\frac{89 \times 100}{0.25}$
$=35600\, g\, mol ^{-1}$
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Classification of Matter
There are two ways of classifying the matter:
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