General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 12 - Solutions - Questions and Problems - Page 528: 12.48


Please see the work below.

Work Step by Step

We know that $S_2=\frac{P_2S_1}{P_1}$ We plug in the known values to obtain: $S_2=\frac{3.74099atm(\frac{0.00175g}{100ml})}{1.00atm}=\frac{0.006546g}{100ml}$
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