Question:
Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve $10 \mathrm{ppm}$ of iron in $100 \mathrm{~kg}$ of wheat is___________ .Atomic weight: $\mathrm{Fe}=55.85 ; \mathrm{S}=32.00 ; \mathrm{O}=16.00$
Solution:
(4.96)
$10=\frac{\text { Mass of Fe }(\text { in } \mathrm{g})}{100 \times 1000} \times 10^{6}$
$\Rightarrow$ Mass of $\mathrm{Fe}=1 \mathrm{~g}$
Molar mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}=278$
$56 \mathrm{~g}$ of iron present in 1 mole of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$
$1 \mathrm{~g}$ of Fe present in $\frac{278}{56} \mathrm{~g}$ of salt $=4.96 \mathrm{~g}$