Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 13 - Solutions - Exercises - Problems - Page 480: 41

Answer

(a) 7.640 % (b) 3.498 % (c) 4.64 %

Work Step by Step

Mass percent = (Mass solute X 100)/ (Mass solute + Mass solvent) (a) 41.2 g $C_{12}H_{22}O_{11}$ in 498 g $H_{2}O$ = (41.2 X 100)/ (498 + 41.2) = 4120/539.2 = 7.640 % (b) 178 mg $C_{6}H_{11}O_{6}$ in 4.91 g $H_{2}O$ = (178 X 10^-3 X 100)/ (0.178 + 4.91) = 17.8 / 5.088 = 3.498 % (c) 7.55 g NaCl in 155 g $H_{2}O$ = (7.55 X 100)/(7.55+155) = 755/162.55 = 4.64 %
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