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) ?

Let the abundance of ¹⁰B be x% and ¹¹B be (100 − x)%. Then the average atomic mass = [10·x + 11·(100 − x)] / 100 = 10·81. Solving: 10x + 1100 − 11x = 1081, which gives −x = −19, so x = 19. Hence ¹⁰B is 19% and ¹¹B is 81%. Therefore the correct answer is 19% and 81%.