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Calculatethe normality of a solution of FeSO_(4).7H_(2)O containing 2.4g/100 mL (Fe=56,S=32,O=16,H=1) which converts to ferric form in a reaction . |
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Answer» Solution :Weight per ML =2.4 g Equivalents/100 mL =`(2.4)/(278) =0.0086 (" eq.wt .of" FeSO_(4).7H_(2)O=278)` `{:{("As "Fe^(2)toFe^(3+)),("eq.wt of "FeSO_(4).7H_(2)O=("MOLECULAR wt")/("change in ON")=278/1):}}`. Thus m.e per 100 mL =`0.0086xx1000=8.6 " "...(Eqn.3)` Normality of solution `=(v.e)/("VOLUME in mL") ""...(Eqn.1)` `=(8.6)/100` `= 0.086` N |
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