Hướng dẫn :

${{n}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}=\frac{{{m}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}}{{{M}_{{{C}_{12}}{{H}_{22}}{{O}_{11}}}}}=\frac{23,94}{342}=0,07mol$

${{C}_{12}}{{H}_{22}}{{O}_{11}}+{{H}_{2}}O\xrightarrow[{{t}^{o}}]{{{H}^{+}}}{{C}_{6}}{{H}_{12}}{{O}_{6}}+{{C}_{6}}{{H}_{12}}{{O}_{6}}$

0,07                        $\to $    0,07 mol

Hiệu suất phản ứng là $80$%$\Rightarrow {{m}_{glucozo}}={{n}_{glucozo}}.{{M}_{glucozo}}.H=0,07.180.0,8=10,08g$

                      $\Rightarrow $ Chọn đáp án A.