Calculate the normality of solution containing 3.15 g of hydrated oxalic acid `(H_2C_2O_4. 2H_2O)` in 250 ml of solution ( Mol. Mass = 126).

Explanation

Mass of oxalic acid = 3.15 g
Equivalent mass of oxalic acid `=( " Mol. mass ")/(" Basicity")`
`=(126)/(2)= 63` g `"equiv"^(-1)`
Equivalent of oxalic acid `=(" Mass of solute ")/( "Eq. Mass ")`
`(3.15 g)/( 63 g " equiv"^(-1))= 0.05 " equiv"^(-1)`
Volume of solution = 250 ml `= (250 )/(1000)L = 0.25 L`
Normality =`("Equivalent of solute")/("Volume of solution in L ")`
`=(0.05 "equiv")/(0.25L)= 0.2 N`