Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 13 - Questions and Problems - Page 598: 13.17

Answer

(a) 0.0610 m sucrose. (b) 2.04 m ethylene glicol.

Work Step by Step

(a) 685 g = 0.685 kg $ C_{12}H_{22}O_{11} $ : ( 1.008 $\times$ 22 )+ ( 12.01 $\times$ 12 )+ ( 16.00 $\times$ 11 )= 342.30 g/mol - Calculate the amount of moles: $$ 14.3 \space g \times \frac{1 \space mol}{ 342.30 \space g} = 0.0418 \space mol$$ - Calculate the molality: $$ \frac{ 0.0418 \space mol}{ 0.685 \space kg} = 0.0610 \space m $$ (b) 3505 g = 3.505 kg - Calculate the molality: $$ \frac{7.15 \space mol}{ 3.505 \space kg} = 2.04 \space m $$
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