A) \[N{{a}_{2}}S{{O}_{3}}\]
B) \[NaHS{{O}_{4}}\]
C) \[N{{a}_{2}}S{{O}_{4}}\]
D) \[N{{a}_{2}}S\]
Correct Answer: A
Solution :
\[N{{a}_{2}}S{{O}_{3}}\]reacts with hot and dil.\[{{H}_{2}}S{{O}_{4}}\]to give \[S{{O}_{2}}\]gas which decolourise bromine water \[N{{a}_{2}}S{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}N{{a}_{2}}S{{O}_{4}}+S{{O}_{2}}+{{H}_{2}}O\] \[B{{r}_{2}}+{{H}_{2}}O\xrightarrow[{}]{{}}2HBr+[O]\] \[S{{O}_{2}}+[O]\xrightarrow[{}]{{}}S{{O}_{3}}\] decolourisation of bromine water.
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