When NaCl is dopped with 1.0 x 10^-3 mole of SrCl₂, the number of cation vacancy is

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$2 \times 6.023 \times 10^{20}$

$6.023 \times 10^{20}$

$6.023 \times 10^{18}$

$3.011 \times 10^{20}$

answer is B.

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Detailed Solution

Two Na+ ions are replaced by one Sr²⁺ ion to maintain electrical neutrality.
Hence, number of vacancies = Number of Sr²⁺ ions doped = 1.0 x 10^-3 x 6.02 x 10²³ = 6.02 x 10²⁰

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