$*)\begin{cases}CH_3COOH:x mol \\ CH_2=CH-COOH:y mol\\C_2H_5COOH:z mol \end{cases}\Rightarrow \begin{cases}60x+72y+74z=3,15 gam \\ y=\frac{3,2}{160} mol\\x+y+z=\frac{90\times 0,5}{1000} \end{cases}$$\Rightarrow \begin{cases}x=0,01 mol \\ y=0,02 mol\\z=0,015 mol \end{cases}\Rightarrow \%CH_3COOH=\frac{74\times 0,015\times 100}{3,15}=35,24\%$
$\Rightarrow Chon:A$