• # question_answer $KMn{{O}_{4}}$ reacts with ferrous sulphate according to the equation $MnO_{4}^{-}+5F{{e}^{2+}}+8{{H}^{+}}\to M{{n}^{2+}}+5F{{e}^{3+}}+4{{H}_{2}}O$ Here $10ml$ of $0.1M$ $KMn{{O}_{4}}$ is equivalent to [CPMT 1999] A) $20ml$ of $0.1M$ $FeS{{O}_{4}}$ B) $30ml$ of $0.1M$ $FeS{{O}_{4}}$ C) $40ml$ of $0.1M$ $FeS{{O}_{4}}$ D) $50ml$ of $0.1M$ $FeS{{O}_{4}}$

In this reaction $\underset{1}{\mathop{MnO_{4}^{-}}}\,+\underset{5}{\mathop{5F{{e}^{2+}}}}\,+8{{H}^{+}}\to M{{n}^{2+}}+5F{{e}^{3+}}+4{{H}_{2}}O$ 5 time quantity of $F{{e}^{2+}}$ consumed. So 5 time of $FeS{{O}_{4}}$ will be equivalent to 50 ml