Question

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

• A. $96$
• B. $98$
• C. $99$
• D. none of the above

Answer 1

The correct option is A $96$

As we know, density $\left(d\right)=\frac{nM}{V{N}_A}$

Here, $a$ is the edge length of the unit cell, $M$ is the molecular mass and $V$ is the volume of an unit cell.
$d=\frac{4\times 24}{(4\times {10}^{-8}{)}^3\times 6\times {10}^{23}}=2.5$ g/cc
The observed density is $2.4$ g/cc.
So, the percentage occupancy of lattice points by element X is $\frac{2.4\times 100}{2.5}=96\%$