\( 1.78 \mathrm{~g} \) 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

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 \(=\frac{100 \times 89}{0.25}=35600 \mathrm{gmol}^{-1}\)