General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

Published by Pearson
ISBN 10: 0321967461
ISBN 13: 978-0-32196-746-6

Chapter 9 - Solutions - 9.4 Solution Concentrations and Reactions - Questions and Problems - Page 348: 9.44

Answer

a. 2.50 M glucose solution. b. 1.00 M HCl solution. c. 2.14 M NaOH solution.

Work Step by Step

a. - Write the molarity expression and evaluate it: $$Molarity(M) = \frac{moles \space of \space solute}{liters \space of \space solution}$$ $$= \frac{0.500 \space mole \space glucose}{ 0.200 \space L \space solution} = 2.50 \space M \space glucose \space solution$$ b. - Calculate or find the molar mass for $ HCl $: $ HCl $ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$ 73.0 \space g \times \frac{1 \space mole}{ 36.46 \space g} = 2.00 \space moles$$ - Write the molarity expression and evaluate it: $$Molarity(M) = \frac{moles \space of \space solute}{liters \space of \space solution}$$ $$= \frac{2.00 \space moles \space HCl}{ 2.00 \space L \space solution} = 1.00 \space M \space HCl \space solution$$ c. - Calculate or find the molar mass for $ NaOH $: $ NaOH $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.00 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$ 30.0 \space g \times \frac{1 \space mole}{ 40.00 \space g} = 0.750 \space mole$$ $1 \space L = 1000 \space mL$ $$Volume(L) = 350. \space mL \times \frac{1 \space L}{1000 \space mL} = 0.350 \space L$$ - Write the molarity expression and evaluate it: $$Molarity(M) = \frac{moles \space of \space solute}{liters \space of \space solution}$$ $$= \frac{0.750 \space mole \space NaOH}{ 0.350 \space L \space solution} = 2.14 \space M \space NaOH \space solution$$
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