Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 11 - Properties of Solutions - Exercises - Page 545: 49


$1.14\times 10^{-3}\frac{mol}{L}$

Work Step by Step

We know that $C=kP$......eq(1) This can be rearranged as: $k=\frac{C}{P}$ We plug in the known values to obtain: $k=\frac{8.21\times 10^{-4}}{0.790}=1.04\times 10^{-3}\frac{mol}{atm.L}$ We plug in the known values in equation (1) to obtain: $C=\frac{1.04\times 10^{-3}mol}{L.atm}\times 1.10atm=1.14\times 10^{-3}\frac{mol}{L}$
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