Fortification of food with iron is done using FeSO 4 × 7 H 2 O . The mass in grams of the FeSO 4 × 7 H 2 O 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 ]

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 4 April, Morning Shift — Question 17

Fortification of food with iron is done using $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ . The mass in grams of the $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is $\_\_\_\_\_$ (Nearest integer) [Given : Molar mass of Fe, S and O respectively are 56, 32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

ppm of $\mathrm{Fe}=\frac{\text { mass of } \mathrm{Fe}}{\text { mass of wheat }} \times 10^{6}$

12=\frac{\mathrm{w}}{150 \times 10^{3}} \times 10^{6}

1.8 \mathrm{gm}=\mathrm{WFe}^{\mathrm{F}}

$\therefore$ mass of $\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}$ required $=\frac{278 \times 1.8}{56}$

$=8.93 \mathrm{gm} \approx 9$ grams

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Morning Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Concentration Terms and Their Interconversion