Chemistry - Solid State Question with Solution | TestHub

The density ($\rho$) of a unit cell is given by the formula:

$$\rho =\frac{Z\times M}{N_A\times a^3}$$

where $Z$ is the number of formula units per unit cell, $M$ is the molar mass, $N_A$ is Avogadro's number, and $a$ is the edge length of the unit cell.

For an NaCl type structure (FCC lattice), $Z=4$.

The distance between two nearest cations (e.g., Na⁺-Na⁺) in an FCC lattice is equal to the face diagonal divided by 2, or $a/\sqrt{2}$.

Given distance between nearest cations $=250\sqrt{2}$ pm.

So, $a/\sqrt{2}=250\sqrt{2}$ pm.

$a=250\sqrt{2}\times \sqrt{2}=250\times 2=500$ pm.

Convert $a$ to cm: $a=500\times 10^{-10}$ cm $=5\times 10^{-8}$ cm.

Now, substitute the values into the density formula:

$M=75$ gm/mol

$N_A=6\times 10^{23}$ mol⁻¹

$a^3=(5\times 10^{-8}cm{)}^3=125\times 10^{-24}c{m}^3$

$$\rho =\frac{4\times 75gm/mol}{6\times 10^{23}mo{l}^{-1}\times 125\times 10^{-24}c{m}^3}$$

$$\rho =\frac{300}{6\times 125\times 10^{-1}}=\frac{300}{750\times 10^{-1}}=\frac{300}{75}=4gm/c{m}^3$$