Given that the abundances of isotopes `"_54Fe` , `"_56Fe` and `"_57Fe` are `5%` ,`90%` and `5%` respectively the atomic mass of Fe is

Options

A. 55.85
B. 55.95
C. 55.75
D. 56.05

Answer

To find the atomic mass of iron (Fe) given the abundances of its isotopes, we will use the formula for the weighted average of the atomic masses based on their abundances. Here are the steps:

### Step-by-Step Solution:

1. **Identify the isotopes and their abundances**:
- Isotope `"_54Fe"` has an abundance of `5%`.
- Isotope `"_56Fe"` has an abundance of `90%`.
- Isotope `"_57Fe"` has an abundance of `5%`.

2. **Convert percentages to decimal form**:
- For calculations, convert the percentages into decimal form:
- `"_54Fe"`: \( \frac{5}{100} = 0.05 \)
- `"_56Fe"`: \( \frac{90}{100} = 0.90 \)
- `"_57Fe"`: \( \frac{5}{100} = 0.05 \)

3. **Identify the atomic masses of the isotopes**:
- The atomic masses (or mass numbers) of the isotopes are:
- \( M_1 = 54 \) (for `"_54Fe"`)
- \( M_2 = 56 \) (for `"_56Fe"`)
- \( M_3 = 57 \) (for `"_57Fe"`)

4. **Use the weighted average formula**:
- The formula for the average atomic mass is:
\[ \text{Average Atomic Mass} = \frac{(M_1 \cdot x_1) + (M_2 \cdot x_2) + (M_3 \cdot x_3)}{x_1 + x_2 + x_3} \]
- Substituting the values:
\[ \text{Average Atomic Mass} = \frac{(54 \cdot 0.05) + (56 \cdot 0.90) + (57 \cdot 0.05)}{0.05 + 0.90 + 0.05} \]

5. **Calculate the numerator**:
- Calculate each term:
- \( 54 \cdot 0.05 = 2.7 \)
- \( 56 \cdot 0.90 = 50.4 \)
- \( 57 \cdot 0.05 = 2.85 \)
- Now add these values together:
\[ 2.7 + 50.4 + 2.85 = 55.95 \]

6. **Calculate the denominator**:
- The denominator is simply the sum of the abundances:
\[ 0.05 + 0.90 + 0.05 = 1.00 \]

7. **Final calculation**:
- Now divide the total from the numerator by the total from the denominator:
\[ \text{Average Atomic Mass} = \frac{55.95}{1.00} = 55.95 \]

### Conclusion:
The atomic mass of iron (Fe) is **55.95**.