Cho m gam hỗn hợp E gồm một peptit X và một peptit Y (biết số nguyên tử

* Hướng dẫn giải

X là tetrapeptit, Y là pentapeptit

$n_{N2} =7,392/22,4=0,33=>n_N =0,66mol$

$E \underset{\text{m}}{\underset{}{\left\{\begin{array}{c}CONH:0,66 \\ {\text{CH}}_2:x \\ {\text{H}}_{2}0:y \end{array}\right.}}\stackrel{+NaOH}{\to } Muoi \underset{\text{m}+23,7}{\underset{}{\left\{\begin{array}{c}COONa:0,66\\ {\text{NH}}_2:0,66\\ {\text{CH}}_2:x\end{array}\right.}}\stackrel{+O_2}{\to }\left\{\begin{array}{c}{\text{Na}}_2{\text{CO}}_3:0,33 \\ \stackrel{BTNT:C}{\to }{\text{CO}}_2:x+0,33 \\ {\text{H}}_{2}0:x+0,66\end{array}\right.$

$mbình tăng = m_{C{O}_2} +m_{H_{2}O} =>44.\left(x+0,33\right)+18.\left(x+0,66\right)=84,06=>x=0,93mol$

$\begin{array}{l}{m}_{muoi }=0,66.67+0,66.16+0,93.14=67,8gam=>m=67,8–23,7=44,1gam\\ m_E =44,1gam=>0,66.43+14x+18y=44,1=>y=0,15\end{array}$

Gọi $n_{Gly-Na} =amol; n_{Ala-Na} =bmol$

$\to \left\{\begin{array}{c}2a+3b=nC=0,66+0,93\\ 97a+111b=67,8\end{array}\right. \to \left\{\begin{array}{c}a=0,39\\ b=0,27\end{array}\right.$

G/s: $n_{X }=u,n_{Y }=v$

$\to \begin{array}{c}u+v=y=0,15\\ 4u+5v={\text{n}}_N=0,66\end{array} \to \left\{\begin{array}{c}u=0,09\\ v=0,06\end{array}\right.$

$\begin{array}{l}X:Gl{y}_{n}Al{a}_{4-n }\left(0,09\right)\\ Y:Gl{y}_{m}Al{a}_{5-m }\left(0,06\right)\end{array}$

$\to 0,09n+0,06m=n_{Gly} =0,39\to 3n+2m=13\left(n<4,m<5\right)$

$\to \left\{\begin{array}{c}n=3\\ m=2\end{array}\right.\to \left\{\begin{array}{c}X:{\text{Gly}}_{3}Ala\\ Y:{\text{Gly}}_2{\text{Ala}}_3\end{array}\right.$ 

$\%m_X = .\frac{0,09.260}{44,1}100\%=53,1\%$

Đáp án cần chọn là: B