1.

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 .

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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