Question

Sodium crystallizes into BCC unit cell. What is the number of unit cells present in $4.6g$ of sodium? [Atomic mass of $Na=23$]

Answer 1

Given: Sodium crystalises into $bcc$ unit cell.

$\therefore z=$ number of atoms per unit cell$=2$.

No. of atoms in $4.6g$ of $Na$ is

$N=n{N}_A=\frac{4.6}{23}\times 6.023\times {10}^{23}=1.2046\times {10}^{23}\text{atoms}$ where $n=\frac{4.6}{23}=\frac{m}{M}.N_A=6.023\times {10}^{23}$

$\therefore$ No. of unit cells is $\frac{N}{z}=\frac{1.2046\times {10}^{23}}{2}=0.6023\times {10}^{23}$