Answer
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 $|90.2-75|/75=20.23\%$
