JEE MAIN - Chemistry (2020 - 8th January Morning Slot - No. 8)
Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to
achieve 10 ppm of iron in 100 kg of wheat is _______.
Atomic weight : Fe = 55.85; S = 32.00; O = 16.00
Atomic weight : Fe = 55.85; S = 32.00; O = 16.00
Answer
4.95to4.97
Explanation
FeSO4.7H2O (M = 277.85)
PPM =
$$ \Rightarrow $$ 10 =
$$ \Rightarrow $$ Mass of Iron = 1 gm
Molecular mass of FeSO4.7H2O is 277.85
55.85 gm iron is present in 277.85 gm of salt
1 gm iron is present in = $${{277.85} \over {55.85}}$$ = 4.97 gm of salt.
PPM =
Mass of Iron
Mass of wheat
$$ \times $$ 106
$$ \Rightarrow $$ 10 =
Mass of Iron
100 $$ \times $$ 103
$$ \times $$ 106
$$ \Rightarrow $$ Mass of Iron = 1 gm
Molecular mass of FeSO4.7H2O is 277.85
55.85 gm iron is present in 277.85 gm of salt
1 gm iron is present in = $${{277.85} \over {55.85}}$$ = 4.97 gm of salt.
Comments (0)
