Calculate the normality of solution containing 6.3g of hydrated oxalic acid in
500 ml of solution
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molecular weight = H2C2O4.2H2O......v.f. = 2
equivalent weight(Ew) = Molecular weight / valancy factor = 126/2 = 63
normality = (wt/Ew) x (1000/V(volume in ml)
normality = (6.3/63) x (1000/500)
normality = 0.1 x 2 = 0.2 N
I hope it will help you
regards
equivalent weight(Ew) = Molecular weight / valancy factor = 126/2 = 63
normality = (wt/Ew) x (1000/V(volume in ml)
normality = (6.3/63) x (1000/500)
normality = 0.1 x 2 = 0.2 N
I hope it will help you
regards
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