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Question

The Mn3O4 formed on strong heating of a sample of MnSO44H2O was dissolved in 100 cm3 of 0.1 N FeSO4 containing dilute H2SO4. The resulting solution reacted completely with 50 cm3 of KMnO4 solution. 25 cm3 of this KMnO4 solution requires 30 cm3 of 0.1 N FeSO4 solution for complete reaction. Calculate the amount of MnSO44H2O in the sample.

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Solution

MnSO44H2OHeat−−Mn3O4
Mn3O4 is dissolved in ferrous sulphate solution and it reduced from Mn(8/3)+ to Mn2+. The excess of FeSO4 is estimated by doing titration with KMnO4 solution. The normality of KMnO4 solution is determined by another ferrous sulphate solution.
For normality of KMnO4 solution:
25×N=30×0.1
N=30×0.125=325
Let the volume of unreacted FeSO4 solution be V mL
V mL of 0.1 N FeSO4=50 mL of 325N KMnO4
or V=5.0×30.1×25=60 mL
Volume of FeSO4 used for Mn3O4=(10060) mL
=40 mL
40 mL of 0.1 N FeSO440 mL of 0.1 N Mn3O4
40 mL of 0.1 N MnSO44H2O
Mass of MnSO44H2O=E×0.1×401000=E250g
Equivalent mass of
MnSO44H2O=M(832)=3M2=3×2232
Mass of MnSO44H2O=3×2232×250=1.338 g.

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