Naturally occurring Boron is made up of 19.9% $^{10}$B and 80.1% $^{11}$B. The relative atomic mass of Boron is

Analyze the question: We are given the percentage abundance of two isotopes of boron and asked to find the relative atomic mass (average atomic mass).

Concept: Relative atomic mass is the weighted average of the isotopic masses:

RAM = (mass₁ × %₁ + mass₂ × %₂) / 100

Solution:

• Contribution of ¹⁰B: 10 × 19.9 = 199
• Contribution of ¹¹B: 11 × 80.1 = 881.1

Sum = 199 + 881.1 = 1080.1

RAM = 1080.1 / 100 = 10.801 ≈ 10.8

Check: The answer must lie between 10 and 11. Since ¹¹B is far more abundant (about 80%), the answer should be much closer to 11, which fits 10.8.

Tip: A quick shortcut is to start at the lighter isotope and add the heavy isotope's fraction: 10 + 0.801 × 1 = 10.801. Also, you can eliminate options A (21.0) and D (10.0) immediately, because the average must fall between the two isotopic masses and be nearer the more abundant one.