Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 5 - Exercises - Page 240d: 59

Answer

a. $$3.54 \times 10^{22} \space molecules $$ b. $$ 1.88 \times 10^{22} \space molecules $$ c. $$ 2.39 \times 10^{21} \space molecules $$

Work Step by Step

a. - Using the molar mass as a conversion factor, find the amount in moles: $$ 1.00 \space g \times \frac{1 \space mole}{ 17.03 \space g} = 0.0587\underline{2} \space mole$$ $$ 0.0587\underline{2} \space mole \space \times \frac{6.02 \times 10^{23} \space molecules }{1 \space mole} = 3.54 \times 10^{22} \space molecules $$ b. - Using the molar mass as a conversion factor, find the amount in moles: $$ 1.00 \space g \times \frac{1 \space mole}{ 32.05 \space g} = 0.0312 \space mole$$ $$ 0.0312 \space mole \space \times \frac{6.02 \times 10^{23} \space molecules }{1 \space mole} = 1.88 \times 10^{22} \space molecules $$ c. - Using the molar mass as a conversion factor, find the amount in moles: $$ 1.00 \space g \times \frac{1 \space mole}{ 252.08 \space g} = 0.00397 \space mole$$ $$ 0.00397 \space mole \space \times \frac{6.02 \times 10^{23} \space molecules }{1 \space mole} = 2.39 \times 10^{21} \space molecules $$
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