Chemistry 9th Edition

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

Chapter 11 - Properties of Solutions - Exercises - Page 547: 75


$2.51\times 10^5\frac{g}{mol}$

Work Step by Step

As we know that $M=\frac{\pi}{RT}$ We plug in the known values to obtain: $M=\frac{0.745torr\times \frac{1atm}{760torr}}{0.08206\frac{L.atm}{mol.K}\times 300K}=3.98\times 10^{-5}\frac{mol}{L}$ Now we can find the molar mass $Molar\space mass=\frac{1.00g}{1.00L\times \frac{3.98\times 10^{-5}}{1L}}=2.51\times 10^5\frac{g}{mol}$
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