Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve <math></math> <math> <mn>10</mn> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">m</mi> </math> of iron in <math><mtext></mtext></math> <math> <mn>100</mn> <mtext> </mtext> <mi mathvariant="normal">k</mi> <mi mathvariant="normal">g</mi> </math> of wheat is Atomic weight : <math></math> <math> <mi mathvariant="normal">F</mi> <mi mathvariant="normal">e</mi> <mo>=</mo> <mn>55.85</mn> <mo>;</mo> <mi mathvariant="normal">S</mi> <mo>=</mo> <mn>32.00</mn> <mo>;</mo> <mi mathvariant="normal">O</mi> <mo>=</mo> <mn>16.00</mn> </math>

496.00 P P M = Mass of Iron Mass of wheat × 10 6 Mass of 10 = Mass of Iron 100 × 10 3 × 10 6 Mole of F e = 1 g m F e = 1 56 contain F e S O 4 7 H 2 O mole of atom = 1 F e in 1 mole ∴ 56 g in 1 g mole 1 56 Mass ∴

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Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve $10 p p m$ of iron in $100 k g$ of wheat is Atomic weight : $F e = 55.85; S = 32.00; O = 16.00$

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Raj Pandey | Contributor-Level 9

a month ago

$496.00$

$P P M = \frac{Mass of Iron}{Mass of wheat} \times 10^{6}$

Mass of $10 = \frac{Mass of Iron}{100 \times 10^{3}} \times 10^{6}$ $\frac{}{^{}}^{}$

Mole of $F e = 1 g m$

$F e = \frac{1}{56}$ contain ${F e S O}_{4} 7 H_{2} O$ $_{}_{}$ mole of atom $= 1$

$F e$ in 1 mole

$∴ 56 g$ in $1 g$ mole

$\frac{}{}$ $\frac{1}{56}$ Mass $∴$

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Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve $10 p p m$ of iron in $100 k g$ of wheat is Atomic weight : $F e = 55.85; S = 32.00; O = 16.00$

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Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 p p m of iron in 100 k g of wheat is Atomic weight : F e = 55.85 ; S = 32.00 ; O = 16.00

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Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve $10 p p m$ of iron in $100 k g$ of wheat is Atomic weight : $F e = 55.85; S = 32.00; O = 16.00$