The concentration of cation vacancies when NaCl doped with 10^-3 mole% of SrCl 2

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Solution

It is given that NaCl is doped with 10^ −3 mol% of SrCl 2 .This means that 100 mol of NaCl is doped with 10 ^−3 mol of SrCl 2 Therefore, 1 mol of NaCl is doped with 10^-3/1000 mol of SrCl2 =10^-5 C ation vacancies produced by one Sr 2+ ion = 1 concentration of the cation vacancies produced by 10^-5 mole of SrCl2 =10^-56.02210^23 =6.022*10^18 per mole

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