If NaCl is doped with 10-3 mole% SrCl2, then the number of cation vacancies will be (NA = 6.022 x 1023 moI-1)

Cationic vacancies = 6.022 × 1023 × 10–5 = 6.022 × 1018

If NaCl is doped with 10-3 mole% SrCl2, then the number of cation vacancies will be (NA = 6.022 x 1023 moI-1)

Cationic vacancies = 6.022 × 1023 × 10–5 = 6.022 × 1018

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If NaCl is doped with 10^-3 mole% SrCl₂, then the number of cation vacancies will be (NA = 6.022 x 10²³ moI^-1)

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6.022 x 10¹⁵ moI^-1

6.022 x 10¹⁸ moI^-1

6.022 x 10²⁰ moI^-1

6.022 x 10¹⁹ moI^-1

answer is B.

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Cationic vacancies = 6.022 × 10²³ × 10^–5
= 6.022 × 10¹⁸

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If NaCl is doped with 10-3 mole% SrCl2, then the number of cation vacancies will be (NA = 6.022 x 1023 moI-1)

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If NaCl is doped with 10^-3 mole% SrCl₂, then the number of cation vacancies will be (NA = 6.022 x 10²³ moI^-1)