Fortification of food with iron is done using $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is _______ (Nearest integer).} (Given : Molar mass of $\mathrm{Fe}, \mathrm{S}$ and O respectively are 56,32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ )
Fortification of food with iron is done using $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is _______ (Nearest integer).} (Given : Molar mass of $\mathrm{Fe}, \mathrm{S}$ and O respectively are 56,32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ )
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Correct Answer: 9
Approach Solution - 1
To determine the mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ needed to achieve 12 ppm of iron in 150 kg of wheat, follow these steps:
Define ppm: 1 ppm is 1 mg of solute per 1 kg of solution. Therefore, for 12 ppm of iron in 150 kg of wheat, the mass of iron needed is:
\( \text{Mass of Fe} = 12 \,\text{mg/kg} \times 150 \,\text{kg} = 1800 \,\text{mg} = 1.8 \,\text{g} \)
To find the mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$, start by calculating its molar mass:
| Element | Atomic Mass (g/mol) | Atoms | Total (g/mol) |
| Fe | 56 | 1 | 56 |
| S | 32 | 1 | 32 |
| O | 16 | 11 (4+7) | 176 |
| H | 1 | 14 (2x7) | 14 |
| Total | 278 g/mol | ||
Calculate the moles of $\mathrm{Fe}$ in 1.8 g:
\( \text{Moles of Fe} = \frac{1.8 \,\text{g}}{56 \,\text{g/mol}} \approx 0.03214 \,\text{mol} \)
Convert moles of $\mathrm{Fe}$ to moles of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ since each unit contains one Fe atom:
\( \text{Moles of } \mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O} = 0.03214 \,\text{mol} \)
Find the mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$:
\( \text{Mass} = 0.03214 \,\text{mol} \times 278 \,\text{g/mol} \approx 8.936 \,\text{g} \)
Round to the nearest integer: 9 g
Verify the solution is within the range 9–9. It is correct.
Therefore, the mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ required is 9 g.
Approach Solution -2
2. Calculate the moles of iron: \[ \text{Moles of iron} = \frac{1.8 \mathrm{~g}}{56 \mathrm{~g/mol}} = 0.0321 \mathrm{~mol} \]
3. Calculate the moles of $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$: \[ \text{Moles of } \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 0.0321 \mathrm{~mol} \]
4. Calculate the mass of $\mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O}$: \[ \text{Molar mass of } \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 56 + 32 + 7 \times 18 = 277 \mathrm{~g/mol} \] \[ \text{Mass of } \mathrm{FeSO}_{4} .7 \mathrm{H}_{2} \mathrm{O} = 0.0321 \mathrm{~mol} \times 277 \mathrm{~g/mol} = 8.8935 \mathrm{~g} \approx 9 \mathrm{~g} \] Therefore, the correct answer is (9).
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