Question

# How many millilitres of 0.02 M KMnO4 solution would be required to exactly titrate 25.0 mL of 0.2 M Fe(NO3)2 solution?

A
50.00 mL
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B
70.00 mL
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C
80.00 mL
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D
None of these
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Solution

## The correct option is A 50.00 mLn factor of KMnO4, Equivalents of KMnO4=5× moles of KMnO4 Normality KMnO4=5× molarity of KMnO4 For Fe(NO3)2, moles and equivalents are equal, as n factor is 1. At the equivalence point, Equivalents of KMnO4 = Equivalents of Fe(NO3)2 or VKMnO4× Normality of KMnO4 =VFe(NO3)2× Normality of Fe(NO3)2 For 0.02 M KMnO4 solution. Normality of KMnO4=(5)(0.02000)=0.1 N and for 0.20 M Fe(NO3)2 solution Normality of Fe(NO3)2=0.20 N putting the values, VKMnO4=(25.00 mL)(0.20000.1000)=50.00 mL

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