Calculated the mass of `FeSO_(4).7H_(2)O`, which must be added in 100 kg of wheat to get 10 PPM of Fe.

Options

A. (a) 2.34 g
B. (b) 3.85 g
C. (c) 4.96 g
D. (d) 6.42 g

Explanation

To solve the problem of calculating the mass of `FeSO4.7H2O` that must be added to 100 kg of wheat to achieve 10 PPM of Fe, we can follow these steps:

### Step 1: Understand the definition of PPM
PPM stands for "parts per million," which means the mass of the solute (in this case, iron) divided by the total mass of the solution (wheat) multiplied by 1,000,000.

### Step 2: Set up the PPM equation
Using the PPM formula:
\[ \text{PPM} = \frac{\text{mass of Fe}}{\text{total mass}} \times 10^6 \]
Given that we want 10 PPM of Fe in 100 kg of wheat, we can substitute the values:
\[ 10 = \frac{\text{mass of Fe}}{100 \text{ kg}} \times 10^6 \]

### Step 3: Convert mass of wheat to grams
Since 1 kg = 1000 grams, we convert 100 kg to grams:
\[ 100 \text{ kg} = 100 \times 1000 = 100,000 \text{ grams} \]

### Step 4: Rearrange the equation to find the mass of Fe
Rearranging the equation to find the mass of Fe:
\[ \text{mass of Fe} = 10 \times \frac{100,000}{10^6} \]
This simplifies to:
\[ \text{mass of Fe} = 10 \times 0.1 = 1 \text{ gram} \]

### Step 5: Calculate the mass of `FeSO4.7H2O` needed
Next, we need to find out how much `FeSO4.7H2O` is required to obtain 1 gram of Fe.

1. The molar mass of `FeSO4.7H2O` is calculated as follows:
- Molar mass of Fe = 56 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol (4 O in SO4)
- Molar mass of H2O = 18 g/mol (7 H2O)

Therefore,
\[ \text{Molar mass of FeSO4.7H2O} = 56 + 32 + (4 \times 16) + (7 \times 18) = 278 \text{ g/mol} \]

2. Since there is 1 gram of Fe in `FeSO4.7H2O`, we can set up a proportion:
\[ \text{If } 278 \text{ g of } FeSO4.7H2O \text{ contains } 56 \text{ g of Fe, then } x \text{ g of } FeSO4.7H2O \text{ contains } 1 \text{ g of Fe.} \]
\[ x = \frac{278 \text{ g} \times 1 \text{ g}}{56 \text{ g}} = 4.9643 \text{ g} \]

### Step 6: Round the answer
Rounding to two decimal places, we find:
\[ x \approx 4.96 \text{ g} \]

### Conclusion
The mass of `FeSO4.7H2O` that must be added to 100 kg of wheat to achieve 10 PPM of Fe is approximately **4.96 grams**.

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