For the reaction, FeS2+KMnO4+H+→Fe3++H2SO4+Mn2++H2O. If the molar mass of FeS2 is M, then equivalent mass of FeS2 would be equal to:
A
M
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B
M10
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C
M11
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D
M15
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Solution
The correct option is DM15 +2−1FeS2→2H2+6SO4++3Fe3+oxidation half reaction n−factor=[1×(3−2)+2(6−(−1)]=15 (∴ n factor of a compound undergoing redox change is equal to no. of moles of electrons lost, gained or exchange by 1 mole of the compound.) So, eq.wt. of FeS2=M15 Hence, (d) is the correct answer.