Question

6.3 g of crystaline oxalic acid present in 100 ml of solution .find molarity and normality...​

Answer 1

Answer:

M = given wt / GmW

= 6.3 / 126 × 1000 / 100

= 0.5

N = M × n - factor

= M × 2

= 0.5 × 2

= 1

Answer 2

Answer:

normality is no. of gram equivalents per unit volume in litres, for oxalic acid, no of gram equivalents will be the given mass divided by equivalent mass.equivalent mass is molecular mass/nfactor.oxalic acid is h2c2o4 so it dissociates as H+ and C2O4^2- ions.hence net positive or negative charge is 2.

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