How many milli gram of iron `(Fe^(2+))` are equal to 1 mL of `0.1055NK_2Cr_2O_7` equivalent?

Options

A. 5.9 mg
B. 0.59 mg
C. 59 mg
D. `5.9 xx 10^(-3) mg`

Answer

To solve the problem, we need to find out how many milligrams of iron (Fe²⁺) are equivalent to 1 mL of 0.1055 N K₂Cr₂O₇.

### Step-by-Step Solution:

1. **Understanding Normality and Equivalent**:
- Normality (N) is a measure of concentration equivalent to molarity multiplied by the number of equivalents per mole. For K₂Cr₂O₇, which acts as an oxidizing agent, its normality indicates how many equivalents of the substance are present in a solution.
- Given: 1 mL of 0.1055 N K₂Cr₂O₇.

2. **Calculate Milliequivalents of K₂Cr₂O₇**:
- The milliequivalents (meq) of K₂Cr₂O₇ in 1 mL can be calculated using the formula:
\[ \text{Milliequivalents} = \text{Volume (mL)} \times \text{Normality (N)} \]
- For 1 mL of 0.1055 N:
\[ \text{Milliequivalents of K₂Cr₂O₇} = 1 \, \text{mL} \times 0.1055 \, \text{N} = 0.1055 \, \text{meq} \]

3. **Determine the n-factor for Fe²⁺**:
- In this redox reaction, Fe²⁺ is oxidized to Fe³⁺, which involves a change in oxidation state of +1. Therefore, the n-factor for Fe²⁺ is 1.

4. **Calculate Equivalent Weight of Fe²⁺**:
- The equivalent weight (EW) of a substance is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{n-factor}} \]
- The molar mass of iron (Fe) is approximately 56 g/mol. Thus:
\[ \text{Equivalent Weight of Fe²⁺} = \frac{56 \, \text{g/mol}}{1} = 56 \, \text{g/equiv} \]

5. **Calculate the Weight of Fe²⁺**:
- The weight of Fe²⁺ that corresponds to the milliequivalents calculated earlier can be found using:
\[ \text{Weight (g)} = \text{Milliequivalents} \times \text{Equivalent Weight} \times \frac{1 \, \text{g}}{1000 \, \text{meq}} \]
- Substituting the values:
\[ \text{Weight (g)} = 0.1055 \, \text{meq} \times 56 \, \text{g/equiv} \times \frac{1}{1000} = 0.0059 \, \text{g} \]

6. **Convert to Milligrams**:
- To convert grams to milligrams:
\[ 0.0059 \, \text{g} = 5.9 \, \text{mg} \]

### Final Answer:
The weight of iron (Fe²⁺) equivalent to 1 mL of 0.1055 N K₂Cr₂O₇ is **5.9 mg**.