Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 3 - The Structure of Crystalline Solids - Questions and Problems - Page 98: 3.7

Answer

$\rho=10.22\frac{g}{cm^3}$; the theoretical value and the experimental value are the same.

Work Step by Step

given ${r}=0.1363 nm$ Atomic weight=95.94g\mol ,and experimental density ${\rho_{exp}}=10.2\frac{g}{ cm^3}$ $\rho=\frac{nA}{{V_{c}}{N_{A}}}$ for BCC, n=2 and $V_{c}=(\frac{4r}{\sqrt{3}})^3$ $N_{A}=6.023\times10^{23} \frac{atom}{mol}$ is Avagadro's number. $\rho=\frac{2\times95.94}{(\frac{4r}{\sqrt{3}})^3\times6.023\times10^{23}}$ $\rho=10.22\frac{g}{cm^3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.