Fortification of food with iron is done using FeSO4 7H2O. The mass in grams of the FeSO4 7H2O required to achieve 12 ppm of iron in 150 kg of wheat is (Nearest integer) [Given: Molar mass of Fe, S and O respectively are 56, 32 and 16 g mol-1]
Answer
9
Solution
12 ppm of iron = 12 g of Fe in 106 g of wheat.
Mass of wheat = 150 kg = 1.5 105 g
⇒ Mass of Fe = 12 1.5 105106 = 1.8 g
Molar mass of FeSO4 7H2O = 56 + 32 + 4(16) + 7(18) = 278 g mol-1
∵ 56 g Fe is present in 278 g of FeSO4 7H2O,
⇒ Mass required = 278 1.856
= 8.94 g 9 g
Answer: 9