A) bleaching powder
B) white vitripl
C) Monies salt
D) microcosmic salt
Correct Answer: C
Solution :
Mohr? s salt \[(FeS{{O}_{4}}.{{(N{{H}_{4}})}_{2}}S{{O}_{4}}.6{{H}_{2}}O)\] \[2KMn{{O}_{4}}+8{{H}_{2}}S{{O}_{4}}+10FeS{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}\] \[+2MnS{{O}_{4}}+5F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+8{{H}_{2}}O\] Hence, Mollys salt decolourises \[\text{KMn}{{\text{O}}_{\text{4}}}\]by reducing \[M{{n}^{7+}}\] ion to \[\text{M}{{\text{n}}^{\text{2+}}}\]ions.
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