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(C) The concentration of iron required is $12 \ ppm$,which means $12 \ g$ of iron per $10^6 \ g$ of wheat.Mass of wheat $= 150 \ kg = 150 	imes 10^3

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) The concentration of iron required is $12 \ ppm$,which means $12 \ g$ of iron per $10^6 \ g$ of wheat.Mass of wheat $= 150 \ kg = 150 	imes 10^3 \ g$.Mass of iron required $(w) = \frac{12}{10^6} 	imes 150 	imes 10^3 = 1.8 \ g$.Molar mass of $FeSO_4 \cdot 7 H_2 O = 56 + 32 + (4 	imes 16) + 7 	imes (2 	imes 1 + 16) = 56 + 32 + 64 + 126 = 278 \ g \ mol^{-1}$.Let the mass of $FeSO_4 \cdot 7 H_2 O$ be $w_1 \ g$.Since $1 \ mol$ of $FeSO_4 \cdot 7 H_2 O$ contains $1 \ mol$ of $Fe$,the moles of $Fe$ are equal.$\frac{w_1}{278} = \frac{1.8}{56}$.$w_1 = \frac{1.8 	imes 278}{56} \approx 8.935 \ g$.Rounding to the nearest integer,we get $9 \ g$.

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