Question

Formula mass of NaCl is $58.45$ g mol${}^{-1}$ and density of its pure form is $2.167$ g cm${}^{-3}$. The average distance between adjacent sodium and chloride ions in the crystal is $2.814\times {10}^{-8}$ cm. Calculate Avogadro's constant.

• A. $0.6\times {10}^{23}$ mol${}^{-1}$
• B. $6.05\times {10}^{23}$ mol${}^{-1}$
• C. $60.5\times {10}^{23}$ mol${}^{-1}$
• D. $0.06\times {10}^{23}$ mol${}^{-1}$

Answer 1

The correct option is B $6.05\times {10}^{23}$ mol${}^{-1}$

As we know, for the given cubic structure, we have
$2r^{+}+2r^{-}=a$
$a=2\times 2.8\mathring{A}$
As we know,
Density $=\frac{nM}{V{N}_o}$m, where $n$ is the number of atoms per unit cell.
Here, $n=4$
$2.17=\frac{4\times 58.5}{5.6\times {10}^{-10}{)}^3\left(N_o\right)}$ $⟹N_o$ $=6.05\times {10}^{23}$ atoms mol${}^{-1}$