Question
$$1.78\,g$$ of an optically active $$L$$ - amino acid $$(A)$$ is treated with $$NaN{O_2}/HCl$$ at $${0^ \circ }C.$$ $$448\,c{m^3}$$ 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
A.
$$36,500\,g\,mo{l^{ - 1}}$$
B.
$$34,500\,g\,mo{l^{ - 1}}$$
C.
$$35,400\,g\,mo{l^{ - 1}}$$
D.
$$35,600\,g\,mo{l^{ - 1}}$$
Answer :
$$35,600\,g\,mo{l^{ - 1}}$$
Solution :
\[L\text{-}\underset{\left( A \right)}{\mathop{\text{amino acid}}}\,\xrightarrow[{{0}^{\circ }}C]{NaN{{O}_{2}}+HCl}{{N}_{2}}\uparrow \]
$$1\,mole\,\,{\text{of}}\,\,L{\text{ - amino acid}}$$ $$ \to {\text{1}}\,mole\,\,{\text{of}}\,\,{N_2}$$ $$ = 22400\,c{m^3}\,\,{\text{of}}\,\,{N_2}\,\,{\text{at }}STP$$
$$448\,c{m^3}\,\,{\text{of}}\,\,{N_2} \equiv 1.78\,g\,\,{\text{of}}$$ $$L{\text{ - amino acid}}$$
$$22,400\,c{m^3}\,\,{\text{of}}\,\,{N_2} \equiv x\,g\,\,{\text{of}}$$ $$L{\text{ - amino acid}}$$
$$x = \frac{{1.78 \times 22,400}}{{448}} = 89\,g\,\,{\text{of}}$$ $$L{\text{ - amino acid}}$$
$${\text{Sample of protein has}}\,\,0.25\% $$ $${\text{amino acid}}{\text{.}}$$
\[\underset{\begin{smallmatrix}
0.25 \\
89
\end{smallmatrix}}{\mathop{\text{Amino acid}}}\,\,\,\,\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix}
100 \\
y
\end{smallmatrix}}{\mathop{\text{Protein}}}\,\]
$$\eqalign{
& y = \frac{{89 \times 100}}{{0.25}} \cr
& \,\,\,\, = 35,600\,g/mol \cr} $$