$n_{NaOH}=0,3 mol= 2n_X$
$\Rightarrow$ X là este phenol.
$n_{H_2O}=0,15 mol$
BTKL, $m_X=29,7+0,15.18-12=20,4g$
$\Rightarrow M_X=\dfrac{20,4}{0,15}=136$
Este có dạng $RCOOC_6H_4R'$
$\Rightarrow R+44+76+R'=136$
$\Leftrightarrow R+R'=16$
$\Rightarrow R=15(CH_3),R'=1(H)$ hoặc $R=1(H), R'=15(CH_3)$
Các CTCT:
$CH_3COOC_6H_5$
$HCOOC_6H_4CH_3$ (o, m, p)
=> 4 CTCT