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5.5 g of a mixutre of FeSO4.7H2O and Fe2...

5.5 g of a mixutre of `FeSO_4.7H_2O` and `Fe_2(SO_4)_3.9H_2O` requires 5.4 " mL of " `0.1 N KMnO_4` solution for complete oxidation. Calculate the number of gram moles of hydrated ferric sulphate in the mixture.

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Equivalent of `KMnO_(4)` = Equivalents of `FeSO_(4).7H_(2)O`
5.4 ml of 0.1 N `KMnO_(4) = 5.4 xx 10^(-4)`equivalents
Amount of `FeSO_(4) = 5.4 xx 10^(-4) xx MoLwt. of FeSO_(4).7H_(2)O = 5.4 xx 10^(-4) xx 278 = 0.150 g`
Total weight of mixture = 5.5 g
Amount of ferric sulphate = 5.5 – 0.150 g = 5.35 g
Hence moles of ferric sulphate = `Mass/MoLwt. = 5.35/562 = 9.5 xx 10^(-3)` gram-mole.
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