A) Bleaching powder
B) White vitriol
C) Mohr's salt
D) Microcosmic salt
E) Laughing gas
Correct Answer: C
Solution :
Mohr?s salt. \[2KMn{{O}_{4}}+8{{H}_{2}}S{{O}_{4}}+10FeS{{O}_{4}}\,\to \]\[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+5F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+8{{H}_{2}}O\] Mohr?s salt decolourises \[KMn{{O}_{4}}\] by reducing \[M{{n}^{+7}}\] ions to \[M{{n}^{+2}}\] ions.
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