18.5 `cm^(3)` of oxalic acid was completely neutralised by `20.0 cm^(3), 0.125` N base. Calculate the (a) normality (b) molarity and ( c) mass of oxalic acid crystals in 1 `dm^(3)` of solution.

Explanation

Normality of aid `=(V xx N)_("base")/("Volume of acid") = (20.0 cm^(3) xx 0.125 N)/(18.5 cm^(3)) = 0.1351` N.
Oxalic acid is a dibasic acid.
`therefore` Molarity of acid `=("Normality of acid")/2 = 0.06757` M .
Mass of oxalic acid crystals in `1 dm^(3)` of solution = normality x equivalent mass = 0.1351 N `xx 63.0` (equivalent mass of oxalic acid crystals) = 63.0 = 8.513 g.