If Nacl is doped with 10^-3mole % of Srcl₂, what is the concentration of cation vacancies?

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Solution

Given,
NaCl is doped with 10^-3 mole % of SrCl₂.This means, 100 moles of NaCl are doped with 10^-3 moles of SrCl₂.
Therefore, 1mole of NaCl is doped with 10^-3 /100 = 10^-5 moles of SrCl₂ or 10^-5 moles of Sr²⁺.
Now,
One Sr²⁺ ion create one cation vacancy.
Therefore, no. of cation vacancies created by 10^-5 moles Sr²⁺
= 10^-5 moles/mole of NaCl
= 10^-5 x 6.022 x 10²³ / mole of NaCl
= 6.022 x 10¹⁸/ mole of NaCl

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