Calculate the normality of 9.0 gram of oxalic acid, `(COOH)_(2)`, dissolved in 250 mL of solution.
Strategy: Find the equivalent mass of oxalic acid by dividing its formula mass by its basicity, which is two, because oxalic acid is a dibasic acid. Use the numerical value of equivalent mass to get gram equivalent mass. Then convert grams of `(COOH)_(2)` to equivalent of `(COOH)_(2)`, which lets to calculate the nirmality

Explanation

Each molecule of oxalic acid contains two acidic hydrogen atoms:
`H-O-overset(O)overset(||)(C)-overset(O)overset(||)(C)-O-H`
Formula mass of `(COOH)_(2)=90u`
Basicity of `(COOH)_(2)=2`
Thus,
Equivalent mass of oxalic acid
`=("Formula mass of" (COOH)_(2))/("Basicity of" (COOH)_(2))`
`=(90u)/2=45 u`
Hence, gram equivalent mass of oxalic acid is `45 g eq.^(-1)`.
Number of equivalents of oxalic acid
`=("Mass of"(COOH)_(2))/("Gram equivalent mass of" (COOH)_(2))`
`=(9 g)/(45 g eq.^(-1))`
`=0.2 eq`.
`"Normality" (N)=(no . eq. (COOH)_(2))/V_(mL)xx(1000 mL)/L`
`=(0.2 eq.)/(250 mL)xx(1000 mL)/L`
`=0.8 eq. L^(-1)` or `0.8 N`