Chemistry - Solid State Question with Solution | TestHub

The formula for the density of a unit cell is given by $\rho =\frac{Z\times M}{N_A\times a^3}$

For an fcc crystal lattice, the number of atoms per unit cell is $Z=4$.

Given:
Molar mass $M=58.5$ g mol${}^{-1}$
Edge length $a=564$ pm $=5.64\times 10^{-8}$ cm
Avogadro's number $N_A=6.022\times 10^{23}$ mol${}^{-1}$

Volume of the unit cell $a^3=(5.64\times 10^{-8}{)}^3\approx 1.794\times 10^{-22}$ cm${}^3$

Substituting the values into the formula:
$\rho =\frac{4\times 58.5}{6.022\times 10^{23}\times 1.794\times 10^{-22}}$

$\rho =\frac{234}{108.03}$

$\rho \approx 2.166$ g/cm${}^3$

Rounding off, we get $2.17$ g/cm${}^3$.

Answer: $2.17$ g/cm${}^3$