Question

At room temperature, sodium crystallises in body-centred cubic lattice with $a=4.24\stackrel{\circ }{A}$. Calculate the theoretical density of sodium. (Atomic mass of $Na=23.0g/mol$)

[Given, ${4.24}^3=76.22$]

• A. $4.004gc{m}^{-3}$
• B. $3.004gc{m}^{-3}$
• C. $1.002gc{m}^{-3}$
• D. $2.002gc{m}^{-3}$

Answer 1

The correct option is C $1.002gc{m}^{-3}$

Sodium crystallizes in body centred lattice.
A body-centred cubic unit cell contains 8 atoms at the 8 corners and 1 at the centre.
Hence,
$\text{Total number of atoms in a unit cell}=8\times \frac{1}{8}+1=2$
Given:
$\text{Edge length, a}=4.24\stackrel{\circ }{A}=4.24\times {10}^{-8}cm$

$\text{Volume of unit cell,}{a}^3=(4.24\times {10}^{-8}{)}^{3}c{m}^3$
So,
$\text{Density,}\rho =\frac{Z\times M}{N_A\times V}$
where,
$Z$ is number of atoms in unit cell
$N_A$ is avagadro number
$M$ is molar mass of element
$V$ is volume of unit cell
$\rho =\frac{2\times 23}{(6.023\times {10}^{23})(4.24\times {10}^{-8}{)}^3}$

$\rho =\frac{2\times 23}{(6.023\times {10}^{23})(76.22\times {10}^{-24})}$

$\rho =1.002gc{m}^{-3}$