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\( 1.78 \mathrm{~g} \) of an optically active \( L \)-amino acid \( (\mathrm{A}) \) is treated with \( \mathrm{N} a \mathrm{~N} \mathrm{O}_{2} / \mathrm{H} \mathrm{Cl} \) at \( 0^{\circ} \mathrm{C} .448 \mathrm{~cm}^{3} \)
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
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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
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The correct answer is:
\( 35,600 \mathrm{~g} \mathrm{~mol}^{-1} \)
Mass of $\mathrm{L}$-amino acid $=178 \mathrm{~g}$
$R-\mathrm{CH}-\mathrm{COOH}-\frac{\mathrm{Na}_{2} / \mathrm{HCl}}{\mathrm{NH}_{2}}$
At $\mathrm{STP}, 1 \mathrm{~mol}=22400 \mathrm{~cm}^{3} \mathrm{C}$
$1 \mathrm{~cm}^{3}=\frac{1}{22400} \mathrm{~mol}$
So, $448 \mathrm{~cm}^{3} \mathrm{~N}_{2}$ is evolved when $\frac{448}{22400} \mathrm{~mol}$ of amino acid reacted
Now, for $\mathrm{L}$-amino acid,
Molar mass of amino acid
$=\frac{22400 \times 1.78 \times 10^{2}}{448}=\frac{178}{2}=89 \mathrm{gmol}^{-1}$
Sample of protein contains $0.25 \%$ amino acid, so $100 \mathrm{~g}$ of protein contain $0.25 \mathrm{~g}$ of amino acid. Therefore,
Molecular mass of protein $=\frac{100 \times 89}{0.25}=35600 \mathrm{gmol}^{-1}$
$R-\mathrm{CH}-\mathrm{COOH}-\frac{\mathrm{Na}_{2} / \mathrm{HCl}}{\mathrm{NH}_{2}}$
At $\mathrm{STP}, 1 \mathrm{~mol}=22400 \mathrm{~cm}^{3} \mathrm{C}$
$1 \mathrm{~cm}^{3}=\frac{1}{22400} \mathrm{~mol}$
So, $448 \mathrm{~cm}^{3} \mathrm{~N}_{2}$ is evolved when $\frac{448}{22400} \mathrm{~mol}$ of amino acid reacted
Now, for $\mathrm{L}$-amino acid,
Molar mass of amino acid
$=\frac{22400 \times 1.78 \times 10^{2}}{448}=\frac{178}{2}=89 \mathrm{gmol}^{-1}$
Sample of protein contains $0.25 \%$ amino acid, so $100 \mathrm{~g}$ of protein contain $0.25 \mathrm{~g}$ of amino acid. Therefore,
Molecular mass of protein $=\frac{100 \times 89}{0.25}=35600 \mathrm{gmol}^{-1}$
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