Lời giải:Đáp án D.
\(\text{BTKL}:{{m}_{\text{E}}}+{{m}_{\text{NaOH}}}={{m}_{\text{muo }\!\!\acute{\mathrm{a}}\!\!\text{ i }}}+{{m}_{\text{ancol }}}+{{m}_{{{\text{H}}_{2}}\text{O}}}\)
\(\rightarrow m_{\mathrm{H}_{2} \mathrm{O}}=0.9 \rightarrow n_{\mathrm{H}_{2} \mathrm{O}}=0,05=n_{\text {peptit }}\)
\(\text{E}\to \left\{ \begin{array}{*{35}{l}}
\text{Val}-\text{Na}:x \\
\text{Gly}-\text{Na}:y \\
\text{Ala}-\text{Na}:0,1 \\
\end{array}\to \left\{ \begin{array}{*{35}{l}}
x+y+0,1=0,44 \\
139x+97y+0,1.111=45,34 \\
\end{array}\to \left\{ \begin{array}{*{35}{l}}
x=0,03 \\
y=0,31 \\
\end{array} \right. \right. \right.\)
\(\rightarrow\left\{\begin{array}{l}
a+b=0,44\left(n_{\mathrm{NaOH}}\right) \\
3,5 a+1,5 b+c+0,05=1,38\left(n_{\mathrm{H}_{2} \mathrm{O}}\right) \\
89 a+57 b+14 c+18.0,05=36\left(m_{\mathrm{E}}\right)
\end{array} \rightarrow\left\{\begin{array}{l}
a=0,16 \\
b=0,28 \\
c=0,35
\end{array}\right.\right.\)
\(\rightarrow n_{\text {ancol }}=a=0,16 \rightarrow M_{\text {ancol }}=46 \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)
Vì \(n_{\text {ancol }}>n_{\text {Val-Na }}, n_{\text {Ala-Na }}\) nên X là este của glyxin: \(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{COOC}_{2} \mathrm{H}_{5}\)
Ta có: \(\frac{n_{\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NO}}}{n_{\text {peptit }}}=\frac{0,28}{0,05}=5,6\) nên Y và Z là pentapeptit và hexapeptit.
\(\left\{ \begin{array}{*{35}{l}}
{{\text{Y}}_{5}}:z \\
{{\text{Z}}_{6}}:t \\
\end{array}\to \left\{ \begin{array}{*{35}{l}}
z+t=0,05 \\
5z+6t=0,28 \\
\end{array}\to \left\{ \begin{array}{*{35}{l}}
z=0,02 \\
t=0,03 \\
\end{array} \right. \right. \right.\)
\({{n}_{{{Z}_{6}}}}={{n}_{\text{Val-Na }}}=0,03\) nên chỉ có Z chứa valin, còn Y thì chỉ chứa glyxin và alanin.
\(\left\{\begin{array}{l}
(\mathrm{Gly})_{n}(\mathrm{Ala})_{(5-n)}: 0,02 \\
(\mathrm{Gly})_{m}(\mathrm{Ala})_{(5-m)} \mathrm{Val}: 0,03
\end{array} \rightarrow 0,02(5-n)+0,03(5-m)=0,1\right.\)
\(\to 2n+3m=15\to \left\{ \begin{array}{*{35}{l}}
n=3 \\
m=3 \\
\end{array} \right.\)
\(\left\{ \begin{align}
& {{(\text{Gly})}_{3}}{{(\text{Ala})}_{2}}:0,02 \\
& {{(\text{Gly})}_{3}}{{(\text{Ala})}_{2}}\text{Val}:0,03 \\
\end{align} \right.\)
\(\to %{{m}_{\text{Y}}}=18,39%\)
