Đáp án đúng: B
1,41
$27,74gX\to \left\{ \begin{array}{l}{{C}_{2}}{{H}_{3}}NO:x\text{ mol}\\C{{H}_{2}}:y\text{ mol}\\{{\text{H}}_{2}}O:z\text{ mol}\end{array} \right.\xrightarrow[{}]{+1,545\text{ mol }{{\text{O}}_{2}}}\left\{ \begin{array}{l}C{{O}_{2}}:\left( 2x+y \right)mol\\{{H}_{2}}O:\left( 1,5x+y+z \right)mol\\{{N}_{2}}:0,5x\text{ mol}\end{array} \right.$
$\Rightarrow \left\{ \begin{array}{l}44.\left( 2x+y \right)-28.0,5x=48,04\\57x+14y+18z=27,74\\\xrightarrow[{}]{BTNT\text{ O}}x+z+2.1,545=2.\left( 2x+y \right)+\left( 1,5x+y+z \right)\end{array} \right.\Rightarrow \left\{ \begin{array}{l}x=0,34\\y=0,52\\z=0,06\end{array} \right.$
=> Số đơn vị aminoaxit trung bình$=\frac{0,34}{0,06}=5,67$
=> Chứng tỏ Y là pentapeptit, Z là hexapeptit.
$\Rightarrow \left\{ \begin{array}{l}5{{n}_{Y}}+6{{n}_{Z}}=0,34\\{{n}_{Y}}+{{n}_{Z}}=0,06\end{array} \right.\Rightarrow \left\{ \begin{array}{l}{{n}_{Y}}=0,02\\{{n}_{Z}}=0,04\end{array} \right.$
Có:$\left\{ \begin{array}{l}{{n}_{Gly}}+{{n}_{Ala}}+{{n}_{Val}}=0,34\text{ mol}\\{{\text{n}}_{C{{O}_{2}}}}=2{{n}_{Gly}}+3{{n}_{Ala}}+5{{n}_{Val}}=1,2\text{ mol}\end{array} \right.\Rightarrow {{n}_{Ala}}+3{{n}_{Val}}=0,52$
$\Rightarrow \frac{0,52-4.0,02}{3}\le {{n}_{Val}}\le \frac{0,52-0,02}{3}\Leftrightarrow 0,147\le {{n}_{Val}}\le 0,167$
=> 3,7$\le $ số đơn vị Val trong Z$\le 4,2\Rightarrow $ Z chứa 4 đơn vị Val =>${{n}_{Val}}=0,12\text{ mol}$
$\Rightarrow \left\{ \begin{array}{l}{{n}_{Gly}}=0,06\text{ mol}\\{{\text{n}}_{Ala}}=0,16\text{ mol}\end{array} \right.\\\Rightarrow \left( a+b \right):c=\left( 97.0,06+111.0,16 \right):\left( 139.0,12 \right)=1,41$
