Question

The edge length of the unit cell of $NaCl$ crystal lattice is $5.623\mathring{A}$, density is $2.16$ g cm${}^{-3}$ and the molar mass of $NaCl$ is $58.5$ g mol${}^{-1}$, the number of moles per unit cell is :

($N_A$ is Avagadro number)

• A. $\frac{4}{N_A}$
• B. $\frac{3}{N_A}$
• C. $\frac{1}{N_A}$
• D. $\frac{2}{N_A}$

Answer 1

Answer: (A)

$\frac{4}{N_A}$

The expression for density is given below:

$d=\frac{zM}{N_A{a}^3}$ $=\frac{z\times 58.5}{{6\times 10}^{23}\times {(5.623\times {10}^{-8})}^3}=2.16$ g cm${}^{-3}$

$⟹z=4$

So, the number of pairs of NaCl per unit cell is $4$.
We know that 1 mole = $N_A$ no. of ions
$\therefore$ 4 NaCl ions =$\frac{4}{N_A}$ moles
Hence, the total no. of moles present per unit cell = $\frac{4}{N_A}$ moles

Option A is correct.