1.78 g of an optically active $L$-amino acid ( $A$ ) is treated with $\mathrm{NaNO}_{2} / \mathrm{HC

Question

1.78 g of an optically active $L$-amino acid ( $A$ ) is treated with $\mathrm{NaNO}_{2} / \mathrm{HCl}$ at $0^{\circ} \mathrm{C} .448 \mathrm{~cm}^{3}$ of nitrogen gas at STP is evolved. A sample of protein has $0.25 \%$ of this amino acid by mass. The molar mass of the protein is\

Professor PC Thomas Classes · Answer

(d) : $L$-amino acid $\xrightarrow[0^{\circ} \mathrm{C}]{\mathrm{NaNO}_{2}+\mathrm{HCl}} \mathrm{N}_{2} \uparrow$\ (A) 1 mole of $L$-amino acid $ightarrow 1$ mole of $\mathrm{N}_{2} =22400 \mathrm{~cm}^{3}$ of $\mathrm{N}_{2}$ at STP\ $448 \mathrm{~cm}^{3}$ of $\mathrm{N}_{2} \equiv 1.78 \mathrm{~g}$ of $L$-amino acid\ $22,400 \mathrm{~cm}^{3}$ of $\mathrm{N}_{2} \equiv x \mathrm{~g}$ of $L$-amino acid\ $x=\frac{1.78 	imes 22,400}{448}=89 \mathrm{~g}$ of $L$-amino acid\ Sample of protein has $0.25 \%$ amino acid.\ $\begin{array}{cc}	ext { Amino acid } & 	ext { Protein } \ 0.25 & 100 \ 89 & y\end{array}$\ $y=\frac{89 	imes 100}{0.25}=35,600 \mathrm{~g} / \mathrm{mol}$