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).

Calculate the normality of solution containing 3.15 g of hydrated oxal

1.	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).
Answer» SOLUTION :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 solutionin L ")` `=(0.05 "equiv")/(0.25L)= 0.2 N`