Đáp án:
$C_2H_5OH$
Giải thích các bước giải:
Gọi CT ancol là $C_nH_{2n+1}OH$
$C_6H_5OH+3Br_2\to C_6H_2(Br)_3OH↓+3HBr$
⇒ $n_{Br_2}=\dfrac{4,8}{160}=0,03\ mol⇒n_{phenol}=0,01\ mol⇒ m_{phenol}=94.0,01 =0,94g\\⇒m_{ancol}=0,184g$
$A+Na$
$C_6H_5OH+Na\to C_6H_5ONa+\dfrac{1}{2}H_2\\0,01\hspace{4cm}0,005\\C_nH_{2n+1}OH+Na\to C_nH_{2n+1}ONa+\dfrac{1}{2}H_2$
$n_{H_2}=\dfrac{0,1568}{22,4}=0,007⇔0,05+\dfrac{1}{2}.n_{C_nH_{2n+1}OH}=0,007 \\⇔ n_{C_nH_{2n+1}OH} = 0,004\ mol⇒ M_{C_nH_{2n+1}OH}=46$
⇒ Ancol : $C_2H_5OH$