Fortification of food with iron is done using $$FeSO_4 \cdot 7H_2O$$. The mass in grams of the $$FeSO_4 \cdot 7H_2O$$ required to achieve 12 ppm of iron in 150 kg of wheat is ________ (Nearest Integer) [Given: Molar mass of Fe, S and O respectively are 56, 32 and 16 g mol$$^{-1}$$]

For fortification, the required concentration is $$12\,\text{ppm}$$ of iron.

Step 1 : Relate ppm to mass.
By definition $$1\,\text{ppm} = 1\,\text{mg}$$ of solute per $$1\,\text{kg}$$ of sample. Therefore $$12\,\text{ppm} = 12\,\text{mg}$$ Fe per $$1\,\text{kg}$$ wheat.

Step 2 : Calculate grams of iron needed for $$150\,\text{kg}$$ wheat.
Required Fe = $$12\,\text{mg kg}^{-1} \times 150\,\text{kg}$$ $$= 1800\,\text{mg} = 1.8\,\text{g}$$

Step 3 : Find molar mass of $$FeSO_4\cdot7H_2O$$.

Molar mass = $$M_{Fe}+M_S+4M_O+7(2M_H+M_O)$$
$$=56 + 32 + 4(16) + 7(2\cdot1 + 16)$$ $$=56 + 32 + 64 + 7(18)$$ $$=56 + 32 + 64 + 126$$ $$=278\,\text{g mol}^{-1}$$

Step 4 : Determine mass fraction of iron in the salt.
Mass fraction of Fe $$= \frac{56}{278}$$ $$\approx 0.2014$$

Step 5 : Compute mass of $$FeSO_4\cdot7H_2O$$ required.
Let $$m$$ be the grams of hydrate needed.
Iron supplied by this mass = $$0.2014\,m$$ (g).
Set equal to required iron: $$0.2014\,m = 1.8$$
$$\Rightarrow m = \frac{1.8}{0.2014}$$ $$\Rightarrow m \approx 8.94\,\text{g}$$

Step 6 : Round to the nearest integer (as asked).
$$m \approx 9\,\text{g}$$

Hence, about 9 grams of $$FeSO_4\cdot7H_2O$$ are needed to fortify 150 kg of wheat to a level of 12 ppm iron.