32g of a sample of `FeSO_(4) .7H_(2)O` were dissolved in dilute sulphuric aid and water and its volue was made up to 1litre. 25mL of this solution required 20mL of `0.02 M KMnO_(4)` solution for complete oxidation. Calculate the mass% of `FeSO_(4) .7 H_(2)O` in the sample.

Options

A. 34.75
B. 69.5
C. 89.5
D. none of these

Answer

To solve the problem step by step, we will follow the outlined procedure to calculate the mass percentage of `FeSO4 . 7H2O` in the sample.

### Step 1: Calculate the number of equivalents of KMnO4 used
The number of equivalents of KMnO4 can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (L)} \times \text{n factor} \]
Given:
- Molarity of KMnO4 = 0.02 M
- Volume of KMnO4 = 20 mL = 0.020 L
- n factor for KMnO4 = 5 (since it changes from +7 to +2, gaining 5 electrons)

Calculating the equivalents:
\[ \text{Number of equivalents of KMnO4} = 0.02 \, \text{mol/L} \times 0.020 \, \text{L} \times 5 = 0.002 \, \text{equivalents} \]

### Step 2: Determine the equivalents of `FeSO4 . 7H2O`
Since the reaction between KMnO4 and `FeSO4 . 7H2O` is a 1:1 reaction in terms of equivalents, the number of equivalents of `FeSO4 . 7H2O` in the 25 mL solution is also 0.002 equivalents.

### Step 3: Calculate the equivalents in 1 liter of solution
To find the total equivalents in 1 liter (1000 mL) of the solution, we can set up a proportion:
\[ \text{Equivalents in 1 L} = \left( \frac{0.002 \, \text{equivalents}}{25 \, \text{mL}} \right) \times 1000 \, \text{mL} = 0.08 \, \text{equivalents} \]

### Step 4: Calculate the number of moles of `FeSO4 . 7H2O`
The number of moles of `FeSO4 . 7H2O` can be calculated using the relation:
\[ \text{Number of moles} = \frac{\text{Number of equivalents}}{\text{n factor}} \]
Given that the n factor for `FeSO4 . 7H2O` is 1:
\[ \text{Number of moles of FeSO4 . 7H2O} = \frac{0.08 \, \text{equivalents}}{1} = 0.08 \, \text{moles} \]

### Step 5: Calculate the mass of `FeSO4 . 7H2O`
To find the mass of `FeSO4 . 7H2O`, we use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
The molar mass of `FeSO4 . 7H2O` is approximately 278.02 g/mol. Thus:
\[ \text{Mass of FeSO4 . 7H2O} = 0.08 \, \text{moles} \times 278.02 \, \text{g/mol} = 22.24 \, \text{g} \]

### Step 6: Calculate the mass percentage of `FeSO4 . 7H2O`
Finally, we can calculate the mass percentage of `FeSO4 . 7H2O` in the sample:
\[ \text{Mass \%} = \left( \frac{\text{Mass of FeSO4 . 7H2O}}{\text{Total mass of sample}} \right) \times 100 \]
Given that the total mass of the sample is 32 g:
\[ \text{Mass \%} = \left( \frac{22.24 \, \text{g}}{32 \, \text{g}} \right) \times 100 \approx 69.5\% \]

### Final Answer:
The mass percentage of `FeSO4 . 7H2O` in the sample is approximately **69.5%**.

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