A solution is prepared by dissolving 0.63 g of oxalic acid in 100 cm^(3) of water. Find the normality of the solution.

A solution is prepared by dissolving 0.63 g of oxalic acid in 100 cm^(

1.	A solution is prepared by dissolving 0.63 g of oxalic acid in 100 cm^(3) of water. Find the normality of the solution.
Answer» Solution :Molecular mass of oxalic acid, `(COOH)_(2) . 2H_(2)O = 126` BASICITY of oxalic acid =2 `therefore` EQUIVALENT mass of oxalic acid `=126/2 = 63` Number of gram equivalents of oxalic acid DISSOLVED `=0.63/63 = 0.01` Volume of the solution `=100 cm^(3) = 100/1000 L = 0.1 L` NORMALITY of the solution `=("No. of g eq. of oxalic acid")/("Volume in litres")` `=0.01/0.1 = 0.1 g eq. L^(-1)`