If NaCl is doped with $10^{-3}$ mole% $SrCl_2$, then what is the concentration of cation vacancies?

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Solution

Doping of NaCl with $10^{- 3}$ mol % $S r C l_{2}$ means that 100 moles of NaCl are doped with $10^{- 3}$ mol of $S r C l_{2}$
$∴ 1$ mole of NaCl is doped with $S r C l_{2} = \frac{10^{- 3}}{100} m o l e = 10^{- 5}$ mole
As each $S r_{+ 2}$ ion creates one cation vacancy, therefore concentration of cation
vacancies = $10^{- 5}$ mol/mol of NaCl = $10^{- 5} \times 6.023 \times 1023 = 6.023 \times 1018 m o l^{- 1}$

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