$S{{O}_{3}}+{{H}_{2}}O\to {{H}_{2}}S{{O}_{4}}$

0,25            $\to $ 0,25           (mol)

$\begin{align}  & {{n}_{S{{O}_{3}}}}=\frac{{{m}_{S{{O}_{3}}}}}{{{M}_{S{{O}_{3}}}}}=\frac{20}{80}=0,25\left( mol \right) \\  &  \\ \end{align}$

${{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=\frac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{{{V}_{{{H}_{2}}S{{O}_{4}}}}}=\frac{0,25}{0,5}=0,5\left( M \right)$

$\Rightarrow $Chọn đáp án A