Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 12 The Solid State - Study Questions - Page 467e: 58


a) 380.4 pm b) 90.2 pm

Work Step by Step

a) Lattice mass: $(40.078|_{Ca}+47.867|_{Ti}+3\times15.999|_O)\div 6.022\times10^{23}=2.257\times10^{-22}\ g/cell$ Lattice volume: $2.257\times10^{-22}\ g/cell\div 4.10\ g/cm^3=5.506\times 10^{-23} cm^3/cell$ Lattice side length: $\sqrt[3]{ 5.506\times 10^{-23} cm^3}=3.804\times10^{-8}\ cm = 380.4\ pm$ b) Midheight plane: $2\times r_{O^{2-}}+2\times r_{Ti^{4+}}=L$ $2\times100+2\times r_{Ti^{4+}}=380.4$ $ r_{Ti^{4+}}=90.2\ pm$ An error of $|90.2-75|/75=20.23\%$
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.