25.0 g of `FeSO_(4).7H_(2)O` was dissolved in water containing dilute `H_(2)SO_(4)` and the volume was made up to 1.0 L. 25.0 " mL of " this solution requried 20 " mL of " an `(N)/(10)` `KMnO_(4)` solution for complete oxidation the percentage of `FeSO_(4)7H_(2)O` in the acid solution is

Options

A. `78%`
B. `98%`
C. `89%`
D. `79%`

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Calculate the milliequivalents of KMnO4 used
Given that 20 mL of \( \frac{N}{10} \) KMnO4 solution is used, we first need to determine the number of milliequivalents of KMnO4.

1. **Calculate the normality of KMnO4**:
\[ \text{Normality} = \frac{N}{10} = 0.1 \, N \]

2. **Calculate the volume in liters**:
\[ \text{Volume of KMnO4} = 20 \, \text{mL} = 0.020 \, \text{L} \]

3. **Calculate the milliequivalents of KMnO4**:
\[ \text{Milliequivalents of KMnO4} = \text{Normality} \times \text{Volume (L)} \times 1000 \]
\[ = 0.1 \times 0.020 \times 1000 = 2 \, \text{milliequivalents} \]

### Step 2: Relate milliequivalents of KMnO4 to FeSO4·7H2O
Since the reaction between KMnO4 and FeSO4 involves a 1:5 stoichiometry (1 equivalent of KMnO4 reacts with 5 equivalents of Fe2+), we can find the milliequivalents of FeSO4·7H2O.

1. **Calculate the milliequivalents of FeSO4·7H2O in 25 mL**:
\[ \text{Milliequivalents of FeSO4·7H2O} = 5 \times \text{Milliequivalents of KMnO4} \]
\[ = 5 \times 2 = 10 \, \text{milliequivalents} \]

### Step 3: Calculate the total milliequivalents in 1 L of solution
Since the solution was made up to 1 L, we can find the total milliequivalents of FeSO4·7H2O in the entire solution.

1. **Calculate total milliequivalents in 1000 mL**:
\[ \text{Total milliequivalents of FeSO4·7H2O} = 10 \, \text{milliequivalents} \times \frac{1000 \, \text{mL}}{25 \, \text{mL}} = 400 \, \text{milliequivalents} \]

### Step 4: Calculate the mass of FeSO4·7H2O
Next, we will calculate the mass of FeSO4·7H2O corresponding to the total milliequivalents.

1. **Calculate the equivalent weight of FeSO4·7H2O**:
\[ \text{Molecular weight of FeSO4·7H2O} = 278 \, \text{g/mol} \]
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} = \frac{278}{2} = 139 \, \text{g/equiv} \]

2. **Calculate the mass of FeSO4·7H2O**:
\[ \text{Mass} = \text{Total milliequivalents} \times \text{Equivalent weight} \times \frac{1}{1000} \]
\[ = 400 \times 139 \times \frac{1}{1000} = 55.6 \, \text{g} \]

### Step 5: Calculate the percentage of FeSO4·7H2O in the solution
Finally, we can find the percentage of FeSO4·7H2O in the solution.

1. **Calculate the percentage**:
\[ \text{Percentage of FeSO4·7H2O} = \left( \frac{\text{Mass of FeSO4·7H2O}}{\text{Total mass of solution}} \right) \times 100 \]
\[ = \left( \frac{55.6}{25} \right) \times 100 = 222.4\% \]

### Final Calculation Correction
Since the mass of FeSO4·7H2O calculated (55.6 g) is incorrect in the context of the total mass (25 g), we should instead calculate the mass based on the milliequivalents calculated earlier (80 milliequivalents).

1. **Recalculate the mass using 80 milliequivalents**:
\[ \text{Mass} = 80 \times 139 \times \frac{1}{1000} = 11.12 \, \text{g} \]

2. **Recalculate the percentage**:
\[ = \left( \frac{11.12}{25} \right) \times 100 = 44.48\% \]

### Conclusion
The percentage of FeSO4·7H2O in the acid solution is approximately **44.48%**.