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 $\mathrm{Fe}, \mathrm{S}$ and O respectively are 56,32 and $16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let mass of iron }=\mathrm{wg} \ & \Rightarrow \frac{\mathrm{w}}{150 	imes 10^3} 	imes 10^6=12 \ & \Rightarrow \mathrm{w}=150 	imes 12 	imes 10^{-3}=1.8 \mathrm{gm} \end{aligned}$ Let mass of $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}=\mathrm{w}_1 \mathrm{gm}$ $\begin{aligned} & \Rightarrow 	ext { Moles of } \mathrm{Fe}=\frac{1.8}{56}=\left(\frac{\mathrm{w}_1}{56+96+7 	imes 18}ight) \ & \Rightarrow \mathrm{w}_1=8.935 \mathrm{gm} \end{aligned}$

Fortification of food with iron is done using $\mathrm{FeSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}$. The mass in…

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