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