Peptit X và Y đều cấu tạp từ 1 loại  no mạch hở ch...

C

- Hướng dẫn giải

X,Y: C_nx­H_2nx-n+2N_nO_n+1: p mol

\(\left\{ {\begin{array}{*{20}{c}}
{pn(1,5x - 0,75) = 1,2285}\\
{pnx = 0,936}\\
{p\left( {nx{\rm{ }} + {\rm{ }}1{\rm{ }}--{\rm{ }}0,5n} \right){\rm{ }} = {\rm{ }}0.873}
\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{pxp = 0,936}\\
{np = 0,234}\\
{p{\rm{ }} = {\rm{ }}0,054}
\end{array}} \right. \Rightarrow } \right.\left\{ {\begin{array}{*{20}{c}}
{x = 4}\\
{{n_x} = 0,018;}\\
{{n_y} = 0,036;}
\end{array}} \right.\)

Y là k peptit

X là t peptit

\(\left\{ {\begin{array}{*{20}{c}}
{\left( {k - 1} \right){\rm{ }}--{\rm{ }}\left( {t - 1} \right){\rm{ }} = {\rm{ }}2}\\
{BTNT{\rm{ }}C:{\rm{ }}4t.0,018{\rm{ }} + {\rm{ }}4k.0,036{\rm{ }} = {\rm{ }}0,936}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{k = 5}\\
{t = 3}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{X:{\rm{ }}{C_{12}}{H_{23}}{N_3}{O_4}\left( {M{\rm{ }} = {\rm{ }}273} \right)}\\
{Y:{\rm{ }}{C_{20}}{H_{37}}{N_5}{O_6}\left( {M{\rm{ }} = {\rm{ }}443} \right)}
\end{array}} \right.\)     

 \(\frac{m}{{153}}(12 + \frac{{23}}{4} - \frac{4}{2}) + \frac{{2m}}{{443}}(20 + \frac{{37}}{4} - \frac{6}{2}) = \frac{{20,13}}{{22,4}} \to m = 5,1\)