A $1.40 g$ sample containing $FeSO X_{4}$ (molar mass $152 g mo l^{- 1}$ ) as the only oxidisable substance is dissolved in dilute acid and requires $25.0 mL$ of $0.0400 M$ $KMnO X_{4}$ . Calculate the mass percent of $FeSO X_{4}$ in the sample (one decimal place).

Numerical answer type, answer in %, tolerance 0.1.

Answer: 54.3 %

Fe2+ = 5 x mol MnO4- = 5 x (0.0250 x 0.0400) = 0.0050 mol. Mass = 0.0050 x 152 = 0.760 g. Percent = 0.760/1.40 x 100 = 54.29, i.e. about 54.3%.