Question

Caesium bromide has $CsCl$ structure $(BCC$ type of lattice$)$. Its density is $4.49gm.cc$. Calculate the side of the unit cell. $[$Molar mass of $Cs=132.09g/mol]$

Answer 1

For $bcc$, $z=2$ & $M=132.09+80=212g{mol}^{-1}$

$S=4.49g/cc$

$\therefore a^3=\frac{Z\times M}{S\times NA}$

$=\frac{2\times 212}{4.49\times 6.022\times {10}^{23}}$

$\therefore a^3=1.568\times {10}^{-22}=0.1568\times {10}^{-21}$

$\therefore a=0.539\times {10}^{-7}m$

$\therefore$ Side of the unit cell $=0.539\times {10}^{-7}m$