Fortification of food with iron is done using $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}$. The mass in grams of the $\mathrm{FeSO}_4$. $7 \mathrm{H}_2 \mathrm{O}$ required to achieve 12 ppm of iron in 150 kg of wheat is ______ (Nearest integer)
[Given: Molar mass of $\mathrm{Fe}, \mathrm{S}$ and and O respectively are 56, 32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
Answer (integer)
9
Solution
<p>Let mass of iron $=\mathrm{w} g$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{\mathrm{w}}{150 \times 10^3} \times 10^6=12 \\
& \Rightarrow \mathrm{w}=150 \times 12 \times 10^{-3}=1.8 \mathrm{gm}
\end{aligned}$$</p>
<p>Let mass of $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}=\mathrm{w}_1 \mathrm{gm}$</p>
<p>$$\begin{aligned}
& \Rightarrow \text { Moles of } \mathrm{Fe}=\frac{1.8}{56}=\left(\frac{\mathrm{w}_1}{56+96+7 \times 18}\right) \\
& \Rightarrow \mathrm{w}_1=8.935 \mathrm{gm}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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