Question

If NaCl is doped with ${10}^{-3}mol\%$ of $SrC{l}_{2^{\prime }},$ each $S{r}^{2+}$ ion replaces two $N{a}^{+}$ ions, but occupies only one lattice point in place of $N{a}^{+}$ ion. This creates one cation vacancy.

Answer 1

Due to addition of strontium chloride $(SrC{l}_{2,}$ each $S{r}^{2+}$ ion replaces two $N{a}^{+}$ ion. This creates one cation vacancy.

Number of moles of cation vacancies in 100 mol of $NaCl={10}^{-3}$

Number of moles of cation vacancies in 1 mole NaCl

$=\frac{{10}^{-3}}{100}={10}^{-5}mol$

Total number of cation vacancies $={10}^{-5}\times N_A$

$=\left({10}^{-5}mol\right)\times (6.022\times {10}^{23}mo{l}^{-1})$

$=6.022\times {10}^{18}$