1.78 g of an optically active L-amino acid ( | vedclass

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(B) Mass of $L$-amino acid $= 1.78 \ g$. The reaction of an amino acid with $NaNO_{2} / HCl$ (Van Slyke method) releases $1 \ mol$ of $N_{2}$ gas per

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$35,600 \ g \ mol^{-1}$

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(B) Mass of $L$-amino acid $= 1.78 \ g$. The reaction of an amino acid with $NaNO_{2} / HCl$ (Van Slyke method) releases $1 \ mol$ of $N_{2}$ gas per $1 \ mol$ of amino acid. At $STP$,$1 \ mol$ of gas occupies $22400 \ cm^{3}$. Moles of $N_{2}$ evolved $= \frac{448 \ cm^{3}}{22400 \ cm^{3} \ mol^{-1}} = 0.02 \ mol$. Since $1 \ mol$ of amino acid produces $1 \ mol$ of $N_{2}$,moles of amino acid $= 0.02 \ mol$. Molar mass of amino acid $= \frac{	ext{Mass}}{	ext{Moles}} = \frac{1.78 \ g}{0.02 \ mol} = 89 \ g \ mol^{-1}$. The protein contains $0.25 \%$ of this amino acid by mass. Let the molar mass of the protein be $M$. $0.25 \% 	ext{ of } M = 89 \ g \ mol^{-1}$. $\frac{0.25}{100} 	imes M = 89$. $M = \frac{89 	imes 100}{0.25} = 35600 \ g \ mol^{-1}$.

$1.78 \ g$ of an optically active $L$-amino acid $(A)$ is treated with $NaNO_{2} / HCl$ at $0^{\circ}C$. $448 \ cm^{3}$ of nitrogen 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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