© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Interpret the meaning of 10 ppm
10 ppm (parts per million) of iron (Fe) means that for every 1 million parts (by mass) of wheat, there are 10 parts (by mass) of iron.
Step 2: Calculate the mass of iron required
The amount of wheat given is 100 kg = 100 × 10³ g = 100,000 g. Using the definition of ppm,
\text{PPM} = \dfrac{\text{Mass of Iron}}{\text{Mass of Wheat}} \times 10^{6}
We have 10 ppm, so
10 = \dfrac{\text{Mass of Iron}}{(100 \times 10^3)} \times 10^{6}
Rearranging,
\text{Mass of Iron} = \dfrac{10 \times (100 \times 10^3)}{10^6} = 1 \text{ g}
Hence, we need 1 g of iron to ensure 10 ppm in 100 kg of wheat.
Step 3: Relate the mass of iron to the mass of FeSO _4 .7H _2 O
We know ferrous sulfate heptahydrate, FeSO _4 .7H _2 O, supplies the iron content. Its molar mass is given as 277.85 g/mol, while iron (Fe) has a molar mass of 55.85 g/mol. From the formula ratio:
55.85 \text{ g of Fe} \longleftrightarrow 277.85 \text{ g of FeSO}_4.7\text{H}_2\text{O}
Therefore, for 1 g of Fe, the corresponding mass of FeSO _4 .7H _2 O will be:
\dfrac{277.85 \text{ g of salt}}{55.85 \text{ g of Fe}} = 4.97 \text{ g of FeSO}_4.7\text{H}_2\text{O}
Step 4: Conclude the required mass of ferrous sulfate heptahydrate
To provide 1 g of iron, we need about 4.97 g of FeSO _4 .7H _2 O. Thus, 4.97 g of ferrous sulfate heptahydrate will achieve 10 ppm of iron in 100 kg of wheat.
