${{n}_{ZnO}}=\frac{{{m}_{ZnO}}}{{{M}_{ZnO}}}=\frac{6,075}{81}=0,075\left( mol \right)$

$ZnO+{{H}_{2}}S{{O}_{4}}\to ZnS{{O}_{4}}+{{H}_{2}}O$

0,075$\to $0,075                                 (mol)

Theo phương trình ta có:

$\begin{align}  & {{n}_{ZnO}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=0,075\left( mol \right) \\ & \Rightarrow {{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=\frac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{{{V}_{{{H}_{2}}S{{O}_{4}}}}}=\frac{0,075}{0,2}=0,375\left( M \right) \\\end{align}$

$\Rightarrow $Chọn đáp án B