Equal quantities of electricity are passed through 3 votameters containing `FeSO_(4),Fe_(2)(SO_(4))_(2) and Fe(NO_(3))_(3)`. Cosider the following statements
(A) The amounts of iron deposited I `FeSO_(4)` and `Fe_(2)(SO_(4))_(3)` are equal
(B) The amount of iron deposited in `Fe(NO_(3))_(2)` is `2//3^(rd)` of the amount deposited in `FeSO_(4)`
(C) The amount of iron deposited in `Fe_(2)(SO_(4))_(3)` and `Fe(NO_(3))_(3)` are equal

Options

A. (A) is correct
B. (B) is correct
C. (C) is correct
D. Both (A) and (B) are correct

Explanation

To solve the problem, we need to analyze the deposition of iron in the three different electrolytes when equal quantities of electricity are passed through each.

### Step-by-Step Solution:

1. **Identify the Electrolytes and Their Formulas:**
- The three electrolytes are:
- \( \text{FeSO}_4 \)
- \( \text{Fe}_2(\text{SO}_4)_3 \)
- \( \text{Fe(NO}_3)_3 \)

2. **Determine the Oxidation States and Equivalent Weights:**
- For \( \text{FeSO}_4 \):
- Iron (Fe) has an oxidation state of +2.
- The equivalent weight (EW) of Fe in \( \text{FeSO}_4 \) is given by:
\[ \text{EW} = \frac{\text{Molar Mass of Fe}}{2} \]
- For \( \text{Fe}_2(\text{SO}_4)_3 \):
- Iron (Fe) has an oxidation state of +3.
- The equivalent weight of Fe in \( \text{Fe}_2(\text{SO}_4)_3 \) is:
\[ \text{EW} = \frac{\text{Molar Mass of Fe}}{3} \]
- For \( \text{Fe(NO}_3)_3 \):
- Iron (Fe) also has an oxidation state of +3.
- The equivalent weight of Fe in \( \text{Fe(NO}_3)_3 \) is:
\[ \text{EW} = \frac{\text{Molar Mass of Fe}}{3} \]

3. **Relate the Amounts of Iron Deposited:**
- When equal quantities of electricity are passed through the solutions, the amount of substance deposited is proportional to the equivalent weight.
- For \( \text{FeSO}_4 \) and \( \text{Fe}_2(\text{SO}_4)_3 \):
- Let \( W_1 \) be the amount of iron deposited from \( \text{FeSO}_4 \) and \( W_2 \) from \( \text{Fe}_2(\text{SO}_4)_3 \).
- Since \( \text{EW}_{\text{FeSO}_4} = \frac{M}{2} \) and \( \text{EW}_{\text{Fe}_2(\text{SO}_4)_3} = \frac{M}{3} \), we can set up the equation:
\[ W_1 \cdot 2 = W_2 \cdot 3 \implies W_1 = \frac{3}{2} W_2 \]
- This means the amount of iron deposited in \( \text{FeSO}_4 \) is greater than that in \( \text{Fe}_2(\text{SO}_4)_3 \).

4. **Compare \( \text{Fe(NO}_3)_3 \) with \( \text{FeSO}_4 \):**
- Let \( W_3 \) be the amount of iron deposited from \( \text{Fe(NO}_3)_3 \).
- Since \( \text{EW}_{\text{Fe(NO}_3)_3} = \frac{M}{3} \), we can relate it to \( W_1 \):
\[ W_1 \cdot 2 = W_3 \cdot 3 \implies W_3 = \frac{2}{3} W_1 \]

5. **Summarize the Findings:**
- From the above analysis:
- The amounts of iron deposited in \( \text{FeSO}_4 \) and \( \text{Fe}_2(\text{SO}_4)_3 \) are not equal.
- The amount of iron deposited in \( \text{Fe(NO}_3)_3 \) is \( \frac{2}{3} \) of the amount deposited in \( \text{FeSO}_4 \).
- The amounts of iron deposited in \( \text{Fe}_2(\text{SO}_4)_3 \) and \( \text{Fe(NO}_3)_3 \) are equal.

### Conclusion:
- **Statement A**: False (the amounts are not equal).
- **Statement B**: True (the amount in \( \text{Fe(NO}_3)_3 \) is \( \frac{2}{3} \) of that in \( \text{FeSO}_4 \)).
- **Statement C**: True (the amounts in \( \text{Fe}_2(\text{SO}_4)_3 \) and \( \text{Fe(NO}_3)_3 \) are equal).