1.50 g of an impure sample of oxalic acid dihydrate, $H X_{2} C X_{2} O X_{4} \cdot 2 H X_{2} O$ (molar mass 126), is dissolved to make 250 mL of solution. A 25.0 mL portion requires 22.0 mL of 0.0200 M $KMnO X_{4}$ in hot acidic medium. Impurities do not react. Find the percentage purity of the sample, to one decimal place.

Numerical answer type, answer in %, tolerance 0.1.

Answer: 92.4 %

mmol

MnO X_{4} X^{-}

22.0 \times 0.0200 = 0.440

; oxalic acid in 25.0 mL =

0.440 \times 2.5 = 1.10

mmol, so 11.0 mmol in 250 mL. Mass =

11.0 \times 126 = 1386

mg = 1.386 g. Purity = 1.386/1.50 x 100 = 92.4%.