Answer:
Boron
Step-by-step explanation:
Step 1: Given data
- Mass of the isotope 1 (m₁): 10.0129 u
- Abundance of the isotope 1 (A₁): 19.78% (0.1978)
- Mass of the isotope 2 (m₂): 11.00931 u
- Abundance of the isotope 2 (A₂): 80.22% (0.8022)
Step 2: Calculate the average atomic mass (m) of the element
We will use the following expression.
m = m₁ × A₁ + m₂ × A₂
m = 10.0129 u × 0.1978 + 11.00931 u × 0.8022
m = 10.81 u
The element with an m = 10.81 u is Boron.