Question

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 $16 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Hint:

Use the molar mass and stoichiometry to calculate the mass of the compound required.

Answer 1

Correct Answer: 9

1. Calculate the mass of iron required: \[ \text{Mass of iron} = \frac{12 \mathrm{~ppm} \times 150 \mathrm{~kg}}{10^6} = 1.8 \mathrm{~g} \]
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).