General Chemistry (4th Edition)

Published by University Science Books
ISBN 10: 1891389602
ISBN 13: 978-1-89138-960-3

Chapter 2 Atoms and Molecules - Problems: 54

Answer

(Isotopic Mass of B-10)(Abundance of B-10) + (Isotopic Mass of B-11) (Abundance of B-11) = Atomic Mass of Boron Suppose x = Abundance of B-10 This implies that Abundance of B-11 = 1-x Therefore (10.013)(x) + (11.009)(1-x) = 10.811 10.013x + 11.009 – 11.009x = 10.811 -0.996x = -0.198 x = 0.1987 1 – x = 1 – 0.1987 = 0.8012 Abundance of B-10 = 19.9% Abundance of B-11 = 80.1%

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