Hướng dẫn:

${{n}_{C{{l}_{2}}}}=\frac{{{V}_{C{{l}_{2}}}}}{22,4}=\frac{2,24}{22,4}=0,1\left( mol \right)$

$C{{l}_{2}}+2NaOH\to NaCl+NaClO+{{H}_{2}}O$

0.1  0,2                                                     (mol)

Theo phương trình ta có:

${{n}_{NaOH}}=2.{{n}_{C{{l}_{2}}}}=2.0,1=0,2\left( mol \right)$

$\Rightarrow {{V}_{NaOH}}=\frac{{{n}_{NaOH}}}{{{C}_{{{M}_{NaOH}}}}}=\frac{0,2}{0,5}=0,4\left( l \right)=400ml$

Chọn đáp án B