Average Atomic Mass of Iron from Isotopic Abundances

To find the average atomic mass of iron ($\text{Fe}$), we use the weighted average of the atomic masses of its isotopes based on their respective natural abundances:

$\text{Average Atomic Mass} = \frac{\sum (A_i \times \%_i)}{100}$

where:

• $A_i$ is the mass number of isotope $i$
• $\%_i$ is the percentage abundance of isotope $i$

Given:

• For $^{54}\text{Fe}$: $A_1 = 54$, $\text{Abundance} = 5\%$
• For $^{56}\text{Fe}$: $A_2 = 56$, $\text{Abundance} = 90\%$
• For $^{57}\text{Fe}$: $A_3 = 57$, $\text{Abundance} = 5\%$

Substituting these values into the formula:
$\text{Average Atomic Mass} = \frac{(54 \times 5) + (56 \times 90) + (57 \times 5)}{100}$

Calculating each term:

• $54 \times 5 = 270$
• $56 \times 90 = 5040$
• $57 \times 5 = 285$

Summing the products:
$\text{Total Mass} = 270 + 5040 + 285 = 5595$

Now, dividing by $100$:
$\text{Average Atomic Mass} = \frac{5595}{100} = 55.95$

Therefore, the atomic mass of $\text{Fe}$ is $55.95$.

Correct Answer: B