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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2025 · 4 April · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

Fortification of food with iron is done using FeSO_4 · 7 H_2 O ....

Fortification of food with iron is done using FeSO4⋅7H2O\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}. The mass in grams of the FeSO4⋅7H2O\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\mathrm{Fe}, \mathrm{S} and O respectively are 56,32 and 16 g mol−116 \mathrm{~g} \mathrm{~mol}^{-1} ]

Correct answer

0

Step-by-step solution

<br/> Let mass of  iron =wg<br>⇒w150×103×106=12<br>⇒w=150×12×10−3=1.8gm<br/>\begin{aligned}<br />& \text{ Let mass\text{ of } iron }=\mathrm{wg} \\<br>& \Rightarrow \frac{\mathrm{w}}{150 \times 10^3} \times 10^6=12 \\<br>& \Rightarrow \mathrm{w}=150 \times 12 \times 10^{-3}=1.8 \mathrm{gm}<br />\end{aligned}
Let mass of FeSO4⋅7H2\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2\mathrm{O}=\mathrm{w}1gm_1 \mathrm{gm}
<br/>⇒ Moles of Fe=1.856=(w156+96+7×18)<br>⇒w1=8.935gm<br/>\begin{aligned}<br />& \Rightarrow \text{ Moles of } \mathrm{Fe}=\frac{1.8}{56}=\left(\frac{\mathrm{w}_1}{56+96+7 \times 18}\right) \\<br>& \Rightarrow \mathrm{w}_1=8.935 \mathrm{gm}<br />\end{aligned}
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