Question

sodium metal crystallizes in a BCC structure. it's unit cell edge length is 420pm. calculate its density. (atomic mass of sodium =23u,NA = 6.022×10 ^23 mol inverse)

Answer 1

Z = 2 since of BCC structure.
M =23
N =6.02 ×10^ 23
a = 420pm = 4.2 × 10^-8
So density = (Z×M) /(N×a) = 2×23/(4.2³ ×6.02) × 10
=460/444.528 =1.034 g/cm³ ans.
Hope this is ur required answer

Answer 2

Answer : The density of the unit cell is, $1.031g/Cm^{3}$

Solution : Given,

Number of atom in unit cell of BCC (Z) = 2

Edge length = $420pm=420\times 10^{-10}cm$ $(1pm=10^{-10}cm)$

Atomic mass of Na (M) = 23 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      ............(1)

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get  the density.

$\rho=\frac{2\times (23g/mol)}{(6.022\times 10^{23}mol^{-1}) \times(420\times 10^{-10}Cm)^3}=1.031g/Cm^{3}$

Therefore, the density of the unit cell is, $1.031g/Cm^{3}$