Cho X, Y, Z là 3 peptit mạch hở (phân tử có số nguyên tử cacbon tương ứng 8,

* Hướng dẫn giải

$\left\{\begin{array}{c}X\left(8C\right)\\ Y\left(9C\right)\\ Z\left(11C\right)\\ T:{\text{C}}_n{H}_{2n}{O}_2\end{array}\right.\stackrel{quy đ0i}{\to }\underset{249,56\left(\text{g}\right)}{\underset{}{\left\{\begin{array}{c}C{\text{H}}_2:x\\ CONH:y\\ {\text{H}}_{2}O:z\\ {\text{O}}_2:T\end{array}\right.}}$

$\begin{array}{l}\left|\begin{array}{c}P1\stackrel{+O_2}{\to }\left\{\begin{array}{c}{\text{CO}}_2:a mol\\ {\text{H}}_{2}O:a-0,11 mol\end{array}\right.\\ P2\stackrel{+\text{NaOH}:\left(\text{y}+\text{t}\right)\text{mol}}{\to }{\text{H}}_{2}O:z mol+{\text{C}}_2{H}_{5}OH:t mol+G \left\{\begin{array}{c}{\text{C}}_n{H}_{2n}{O}_{2}NNa\\ {\text{C}}_m{H}_{2m-1}{O}_{2}Na\end{array}\right.\end{array}\right.\\ \text{G }\left\{\begin{array}{c}{\text{C}}_n{H}_{2n}{O}_{2}NNa\\ {\text{C}}_m{H}_{2m-1}{O}_{2}Na\end{array}\right.\stackrel{+O_2: 3,385 mol}{\to }\left\{\begin{array}{c}{\text{Na}}_2{\text{CO}}_3:0,5y+0,5t\\ {\text{CO}}_2:a+0,5y-2,5t\\ {\text{H}}_{2}O:x+y-2,5t\end{array}\right.\end{array}$

P1: $n_{CO2} –n_{H2O} =0,11=>0,5y–z=0,11\left(1\right)$

P2:$n_{NaOH} =y+t=>n_{Na2CO3} =0,5y+0,5t$

BT “C”: $x + y -2t = 0,5y + 0,5t + nCO2$

=> $nCO2 = x + 0,5y – 2,5t$

BT “$H”:\left(2x+y+2z\right) + \left(y+t\right)-2z-6t=2n_{H2O}$

$=>n_{H2O }=x+y–2,5t$

BT “O”: $nO\left(muối\right) + 3,385.2 = 3(0,5y + 0,5t) + 2(x + 0,5y – 2,5t) + (x + y – 2,5t)$

=> 3x + 1,5y – 8t = 6,77 (2)

Phương trình về KL: 14x + 43y + 18z + 32t = 249,56:2 (3)

BTKL: 124,78 + 4(y+t) = 133,18 + 18z + 46t

=> 40y – 18z – 6t = 8,4 (4)

(1) (2) (3) (4) => x = 4,98; y = 0,42; z = 0,1; t = 1,1

$\to 4,3 < nC\left(este\right) < 4,6 \to < C\left(este\right) <$

$\to Este: C_4{H}_8{O}_2$

→ $\left\{\begin{array}{c}{\text{n}}_{\text{C}\left(\text{peptit}\right)}=5,4-4,4=1\\ {\text{n}}_{\text{peptit}}=z=0,1\\ {\text{n}}_{\text{CONH}}=0,42\to so mat xích tb=4,2\end{array}\right.$

$\left\{\begin{array}{c}Z:{\text{Gly}}_{4}Ala\left(m\right)\\ Y:{\text{Gly}}_{3}Ala\left(n\right)\\ X:ValAla\left(p\right)\end{array}\right.\to \left\{\begin{array}{c}m+n+p=0,1\\ 11m+9n+8p=1\\ 5m+4n+2p=0,42\end{array}\right.$

→ $\left\{\begin{array}{c}m=0,06\\ n=0,02\\ z=0,02\end{array}\right.$

$\to \%{m_Y}_{ }= .100\% = 4,17\%$

Đáp án cần chọn là: B