Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems: 3.30

Answer

First of all, we need to find out the mass of water: $Mass \space of \space water=2.56mL\times \frac{1.00g}{1.00mL}=2.56g$ Now, we can determine the number of moles as follows: $2.56g\space H_2O\times \frac{1mol\space H_2O}{18.016g\space H_2O}\times \frac{6.022\times 10^{23}molecules\space H_2O}{1mol\space H_2O}=8.56\times 10^{22}moleules$

Work Step by Step

First of all, we need to find out the mass of water: $Mass \space of \space water=2.56mL\times \frac{1.00g}{1.00mL}=2.56g$ Now, we can determine the number of moles as follows: $2.56g\space H_2O\times \frac{1mol\space H_2O}{18.016g\space H_2O}\times \frac{6.022\times 10^{23}molecules\space H_2O}{1mol\space H_2O}=8.56\times 10^{22}moleules$
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