A) \[NaCl\]with\[{{H}_{2\,}}O\]
B) \[N{{H}_{4}}Cl\]with\[NaOH\]
C) \[C{{l}_{2}}\]with cold and dilute\[NaOH\]
D) \[C{{l}_{2}}\]with hot and concentrated\[NaOH\]
Correct Answer: B
Solution :
\[NaOCl\]is used as a bleaching agent and sterilising agent. It is formed by\[C{{l}_{2}}\]with cold and dilute\[NaOH\]. \[NaOH+C{{l}_{2}}\xrightarrow[{}]{{}}NaCl+NaOCl+{{H}_{2}}O\] Cold and dil. \[6NaOH+3C{{l}_{2}}\xrightarrow{{}}5NaCl+NaCl{{O}_{3}}\] Hot and cone. \[+3{{H}_{2}}O\]
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