The relative atomic mass of boron (which exists in two isotopic forms ¹⁰B and ¹¹B) is 10.81. What will be the abundance of ¹⁰B and ¹¹B, respectively (consider a sample of 100 atoms)?

The correct answer is A, 19% and 81%. Only this split of boron-10 and boron-11 gives an average atomic mass of 10.81.

The relative atomic mass of an element is the average of its isotopes' masses, each weighted by how common it is. In 100 atoms, let x be boron-10, so 100 − x are boron-11. Then 10x + 11(100 − x) = 1081, which gives 1100 − x = 1081, so x = 19. That means 19 atoms of boron-10 and 81 of boron-11, or 19% and 81%. A quick check: the average 10.81 is much nearer 11 than 10, so the heavier isotope must be the more common one.

B is wrong because it swaps the two values; 81% of boron-10 would give an average of 10.19.

C is wrong because 38% and 62% give an average of 10.62.

D is wrong because 62% and 38% give an average of 10.38.

Exam tip: average atomic mass = Σ (isotope mass × fraction); it always lies nearer the more abundant isotope.