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: 56

Answer

(Isotopic Mass of Eu-151)(Abundance of Eu-151) + (Isotopic Mass of Eu-153) (Abundance of Eu-153) = Atomic Mass of Eu Suppose x = Abundance of Eu-151 This implies that Abundance of Eu-153= 1-x Therefore (150.9199)(x) + (152.9212)(1-x) = 151.964 150.9199x + 152.9212 – 152.9212x = 151.964 2.0013x = -0.9572 x = 0.4783 1 – x = 1 – 0.4783 = 0.5217 Abundance of Eu-151= 47.83 % Abundance of Eu-153= 52.17%

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