Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 13 - Problems: 13.102

Answer

11kg

Work Step by Step

(change in) T = - K(f) x m Convert -12 degrees F to degrees C: (-12 - 32) x (5 / 9) = -24.4 degrees C (change in) T = -24.4 - 0 = -24.4 degrees C [remember 0 is the normal freezing point of water] K(f) = 1.86 degrees C / molality [can be found on freezing point depression chart] Since you have T and K(f) solve for m: m = (-24.4 degrees C) / ( -1.86 degrees C / m) m =13.118 mol C2H6O2 / 1 kg water This number is per 1 kg of water, since we have 14.5kg we multiply by 14.5 13.118 mol C2H6O2/ 1 kg x 14.5 kg = 190.211 mol C2H6O2 needed convert mol to grams 190.211 mol x (62.08 g / mol) = 11808.298 g or 11.808 kg
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