JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Some Basic Concepts of Chemistry

Ten parts per million of iron in $100\text{ kg}$ of wheat means the wheat must hold $10\times 10^{-6}\times 100\text{ kg}=10^{-3}\text{ kg}=1\text{ g}$ of iron. The moles of iron required are therefore $\frac{1}{55.85}\approx 0.0179\text{ mol}$. Each formula unit of ferrous sulphate heptahydrate, $\mathrm{FeS}{\mathrm{O}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}$, carries exactly one iron atom, and its molar mass is $55.85+32+4\times 16+7\times 18=277.85{\text{ g mol}}^{-1}$. Multiplying the moles of iron by this molar mass gives the mass of salt, $0.0179\times 277.85\approx 4.96\text{ g}$, so a little under five grams of the salt is needed.