JEE Chemistry 2025 Question with Solution

Given: Iron fortification is done using $\mathrm{FeSO}_{4}\cdot 7\mathrm{H}_{2}\mathrm{O}$. Required iron concentration is $12\,\text{ppm}$ in $150\,\text{kg}$ of wheat.

Find: Mass of $\mathrm{FeSO}_{4}\cdot 7\mathrm{H}_{2}\mathrm{O}$ required.

Using ppm, the mass of iron needed is

$$\text{Mass of Fe} = 12\,\text{mg/kg} \times 150\,\text{kg} = 1800\,\text{mg} = 1.8\,\text{g}$$

Now calculate moles of iron:

$$\text{Moles of Fe} = \frac{1.8\,\text{g}}{56\,\text{g/mol}} \approx 0.03214\,\text{mol}$$

Each mole of $\mathrm{FeSO}_{4}\cdot 7\mathrm{H}_{2}\mathrm{O}$ contains one mole of $\mathrm{Fe}$, so

$$\text{Moles of } \mathrm{FeSO}_{4}\cdot 7\mathrm{H}_{2}\mathrm{O} = 0.03214\,\text{mol}$$

Its molar mass from the extracted working is

$$56 + 32 + 11\times 16 + 14\times 1 = 278\,\text{g/mol}$$

Therefore,

$$\text{Mass} = 0.03214\,\text{mol} \times 278\,\text{g/mol} \approx 8.936\,\text{g}$$

Rounding to the nearest integer gives $9\,\text{g}$.

Therefore, the required mass is $9\,\text{g}$, so the numerical answer is 9.

Given: Required iron is $12\,\text{ppm}$ in $150\,\text{kg}$ wheat.

Find: Mass of $\mathrm{FeSO}_{4}\cdot 7\mathrm{H}_{2}\mathrm{O}$.

First find iron required:

$$\text{Mass of Fe} = \frac{12\times 150}{10^6}\,\text{kg} = 1.8\,\text{g}$$

Using the mass fraction of iron in $\mathrm{FeSO}_{4}\cdot 7\mathrm{H}_{2}\mathrm{O}$,

$$\text{Mass of salt} = 1.8 \times \frac{278}{56}$$ $$\text{Mass of salt} \approx 8.94\,\text{g} \approx 9\,\text{g}$$

This shortcut works because one formula unit of the salt contains exactly one iron atom, so the mole ratio between iron and the salt is $1:1$.

Therefore, the numerical answer is 9.