Given that the abundacne of isotopes `.^(54)Fe`, `.^(56)Fe`, and `.^(57)Fe` is 5%, 90% and 5% respectively. The atomic mass of `Fe` is

Options

A. `55.85`
B. `55.95`
C. `55.75`
D. `55.05`

Explanation

To calculate the atomic mass of iron (Fe) given the abundances and atomic masses of its isotopes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Isotopes and Their Abundances**:
- Isotope: \( ^{54}Fe \) with abundance = 5%
- Isotope: \( ^{56}Fe \) with abundance = 90%
- Isotope: \( ^{57}Fe \) with abundance = 5%

2. **Identify the Atomic Masses of the Isotopes**:
- Atomic mass of \( ^{54}Fe \) = 54 u
- Atomic mass of \( ^{56}Fe \) = 56 u
- Atomic mass of \( ^{57}Fe \) = 57 u

3. **Calculate the Weighted Average**:
The average atomic mass can be calculated using the formula:
\[ \text{Average Atomic Mass} = \frac{(abundance_1 \times mass_1) + (abundance_2 \times mass_2) + (abundance_3 \times mass_3)}{100} \]

Plugging in the values:
\[ \text{Average Atomic Mass} = \frac{(5 \times 54) + (90 \times 56) + (5 \times 57)}{100} \]

4. **Perform the Calculations**:
- Calculate each term:
- \( 5 \times 54 = 270 \)
- \( 90 \times 56 = 5040 \)
- \( 5 \times 57 = 285 \)

- Sum the results:
\[ 270 + 5040 + 285 = 5595 \]

5. **Divide by 100**:
\[ \text{Average Atomic Mass} = \frac{5595}{100} = 55.95 \, u \]

### Final Answer:
The atomic mass of iron (Fe) is approximately **55.95 u**.

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