If NaCl is doped with 10^–4 mol % SrCl₂, the concentration of cation vacancies will be $(N_{A} = 6.02 \times 10^{23} m o l^{- 1})$

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$6.02 \times 10^{14} m o l^{- 1}$

$6.02 \times 10^{17} m o l^{- 1}$

$6.02 \times 10^{15} m o l^{- 1}$

$6.02 \times 10^{16} m o l^{- 1}$

answer is D.

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Detailed Solution

One mole SrCl₂ causes one mole cation vacancy.

So, 10^–4% mole SrCl₂ will cause $\frac{10^{- 4}}{100}$

mole cation vacancy.

∴ Number of cation vacancy

$\frac{10^{- 4}}{100} \times N_{A} = 6.02 \times 10^{23 - 6} = 6.02 \times 10^{17} m o l^{- 1}$

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