Hòa tan một lượng ancol X vào nước thu được 6,4 gam dung dịch Y

* Hướng dẫn giải

$$\begin{gathered} \left\{\begin{array}{l}{\mathrm{m}}_{\mathrm{R}\left(\mathrm{OH}\right)\mathrm{m}} = 6,4. 71,875\% = 4,6\\ {\mathrm{m}}_{\mathrm{H}2\mathrm{O} }= 6,4 - 4,6 = 1,8\end{array} \right. \\ \to \left\{\begin{array}{l}{\mathrm{m}}_{\mathrm{R}\left(\mathrm{OH}\right)2} =\frac{4,6}{{\mathrm{M}}_{\mathrm{R}\left(\mathrm{OH}\right)\mathrm{m}}}\\ {\mathrm{n}}_{\mathrm{H}2\mathrm{O} }= 1,8\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{R}{\left(\mathrm{OH}\right)}_{\mathrm{n}}\\ \mathrm{HOH}\end{array}\right.\stackrel{\mathrm{Na}}{\to }\left\{\begin{array}{l}\mathrm{R}{\left(\mathrm{ONa}\right)}_{\mathrm{n}}\\ \mathrm{NaOH}\end{array} + {\mathrm{H}}_2\right. \\ \mathrm{BTNT} \mathrm{H}: \frac{4,6}{{\mathrm{M}}_{\mathrm{R}\left(\mathrm{OH}\right)\mathrm{m}}} + 0,1 = 0,125.2 \\ \to \frac{{\mathrm{M}}_{\mathrm{R}\left(\mathrm{OH}\right)\mathrm{m}}}{\mathrm{n}} = \frac{92}{3} \\ \Rightarrow \left\{\begin{array}{l}\mathrm{n} = 3, {\mathrm{M}}_{\mathrm{R}\left(\mathrm{OH}\right)\mathrm{m}} = 92\\ \mathrm{R}\left(\mathrm{OH}\right)\mathrm{m} \mathrm{là} {\mathrm{C}}_3{\mathrm{H}}_5{\left(\mathrm{OH}\right)}_3\end{array}\right. \\ \mathrm{Vậy} \mathrm{ancol} \mathrm{có} 8\mathrm{H} \Rightarrow \mathrm{Đáp} \mathrm{án} \mathrm{D} \end{gathered}$$