Question

An element X (atomic weight $=24$ g/mol) forms a face centered cubic lattice. If the edge length of the lattice is $4\times {10}^{-8}$ cm and the observed density is $2.40\times {10}^3$ g cm${}^{-3}$, then the percentage occupancy of lattice points by element X is : (Use $N_A=6\times {10}^{23}$)

Answer 1

Given that,

Atomic weight of element $X=24$ $g/mol$

And element $X$ is in FCC.

We know that,

Density of unit cell $=\frac{Z\cdot M}{a^3\times N_A}$

where, $Z=$ Number of atoms of elements in unit cell.

$M=$ Molecular weight

$a^3=$ Volume of unit cell

$N_A=$ Avogadro's Number

$a=$ Edge length

In FCC $H-$ atom is present.

So, Density $=\frac{4\times 24}{64\times 6\times {10}^{23}\times {10}^{-24}}$ $g/c{m}^3=0.25\times {10}^1$ $g/c{m}^3$

Calculated Density $=2.5$ $g/c{m}^3$

$Percentageoccupancy=\frac{Observeddensity}{Calculateddensity}=\frac{2.4}{2.5}\times 100=96\%$