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Q.
If $NaCl$ is doped with $10^{-3}$ mole $\%$ of $SrCl_2$, then number of cationic vacancies is
The Solid State
Solution:
Due to the addition of $SrCl_2$, each $Sr^{2+}$ ion replaces two $Na^+$ ions, but occupies one $Na^+$ lattice point. Thus, this exchange of $Na^+$ ion by $Sr^{2+}$ ion makes one cationic vacancy.
$SrCl_2$ doped $= 10^{-3}$ mol per $100$ mol
$= 10^{-5}$ mol per $1$ mol
$\therefore $ Cation vacancies $= 10^{-5}$ mol per $1$ mol
$= 10^{-5} \times N_0\, mol^{-1}$
$= 10^{-5} \times 6.02 \times 10^{23}$
Total $= 6.02 \times 10^{18}$ cationic vacancies $mol^{-1}$