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InterviewSolution
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In NaCl is doped with `1 xx 10^(-3)` mol percent of `SrCl_(2)` , what is the concentration of cation vacancies ? |
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Answer» The addition of `SrCl_(2)` to NaCl , each `Sr^(+2)` ion replaces two `Na^(+)` ions and occupies only one lattice point in place of `Na^(+)` . Due to this one cation vacancy arised . The number of moles of cation vacancies in 100 moles of NaCl = `1 xx 10^(-3)` The number of moles of cation vacancies in 1 mole NaCl = `(1 xx 10^(3))/(100) = 10^(-5)` mole Total number of cation vacancies = `10^(-5) xx 6.023 xx 10^(23) = 6.023 xx 10^(18)`. |
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