Hỗn hợp E gồm 3 chất: X (là este của amino axit); Y và Z là hai

Đáp án D.

$$\begin{gathered} \text{BTKL}:m_{\text{E}}+m_{\text{NaOH}}=m_{\text{muối }}+m_{\text{ancol }}+m_{{\text{H}}_2\text{O}} \\ \to m_{{\text{H}}_2\text{O}}=0.9\to n_{{\text{H}}_2\text{O}}=0,05=n_{\text{peptit }} \\ \text{E}\stackrel{\text{NaOH}}{\to }\left\{\begin{array}{l}\text{Val}-\text{Na}:x\\ \text{Gly}-\text{Na}:y\\ \text{Ala}-\text{Na}:0,1\end{array}\to \left\{\begin{array}{l}x+y+0,1=0,44\\ 139x+97y+0,1.111=45,34\end{array}\to \left\{\begin{array}{l}x=0,03\\ y=0,31\end{array}\right.\right.\right. \\ \text{E}\left\{\begin{array}{l}{\text{NH}}_2{\text{CH}}_2{\text{COOCH}}_3:a\\ {\text{C}}_2{\text{H}}_3\text{NO}:b\\ {\text{CH}}_2:c\\ {\text{H}}_2\text{O}:0,05\end{array}\right.\to \left\{\begin{array}{l}a+b=0,44\left(n_{\text{NaOH}}\right)\\ 3,5a+1,5b+c+0,05=1,38\left(n_{{\text{H}}_2\text{O}}\right)\\ 89a+57b+14c+18.0,05=36\left(m_{\text{E}}\right)\end{array}\to \left\{\begin{array}{l}a=0,16\\ b=0,28\\ c=0,35\end{array}\right.\right. \\ \to n_{\text{ancol }}=a=0,16\to M_{\text{ancol }}=46\to {\text{C}}_2{\text{H}}_5\text{OH} \end{gathered}$$

Vì $n_{\text{ancol }}>n_{\text{Val-Na }},n_{\text{Ala-Na }}$  nên Y và Z là este của glyxin: ${\text{NH}}_2{\text{CH}}_2{\text{COOC}}_2{\text{H}}_5$

Ta có: $\frac{n_{{\text{C}}_2{\text{H}}_3\text{NO}}}{n_{\text{peptit }}}=\frac{0,28}{0,05}=5,6$ nên Y và Z là pentapeptit và hexapeptit.

$\left\{\begin{array}{l}{\text{Y}}_5:z\\ {\text{Z}}_6:t\end{array}\to \left\{\begin{array}{l}z+t=0,05\\ 5z+6t=0,28\end{array}\to \left\{\begin{array}{l}z=0,02\\ t=0,03\end{array}\right.\right.\right.$

$n_{Z_6}=n_{\text{Val-Na }}=0,03$ nên chỉ có Z chứa valin, còn Y thì chỉ chứa glyxin và alanin.

$$\begin{gathered} \left\{\begin{array}{l}{\left(\text{Gly}\right)}_n{\left(\text{Ala}\right)}_{(5-n)}:0,02\\ {\left(\text{Gly}\right)}_m{\left(\text{Ala}\right)}_{(5-m)}\text{Val}:0,03\end{array}\to 0,02(5-n)+0,03(5-m)=0,1\right. \\ \to 2n+3m=15\to \left\{\begin{array}{l}n=3\\ m=3\end{array}\right. \\ \left\{\begin{array}{l}{\left(Gly\right)}_3{\left(Ala\right)}_2:0,02\\ {\left(Gly\right)}_3{\left(Ala\right)}_{2}Val:0,03\end{array}\right. \\ \to \%m_{\text{Y}}=18,39\% \end{gathered}$$

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