Question

Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is
(Atomic weight: Fe=55.85; S=32.00; O=16.00)

Answer 1

10 ppm of Fe means 10 g of Fe in ${10}^6$ g of wheat. So for 100 kg i.e, ${10}^5$ g of wheat. Fe needed is 1g. So, for 1 g of Fe , the mass of $FeS{O}_4.7H_{2}O$ required is $\frac{278}{56}=4.96\text{g}$