A metal (atomic mass =50) has a body-centred cubic lattice.The density of the metal is $5.91gcm^{-3}$.Find out the volume of the unit cell.$[N_A=6.023\times 10^{23}]$

Question

$\begin{array}{1 1}2.809\times 10^{-23}cm^3\\1.809\times 10^{-23}cm^3\\0.809\times 10^{-23}cm^3\\4.809\times 10^{-23}cm^3\end{array} $

Answer 1

$M=50gmol^{-1}$

$z=2$

$d=5.91gcm^{-3}$

$V=a^3$

$N_A=6.023\times 10^{23}$

$d=\large\frac{z\times M}{N_A\times a^3}$

$a^3=\large\frac{z\times M}{N_A\times d}$

$\;\;\;\;=\large\frac{2\times 50}{5.91\times 6.023\times 10^{23}}$

$\;\;\;\;=2.809\times 10^{-23}cm^3$