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If NaCl is doped with 10^(-3) mol% GaCl(...

If `NaCl` is doped with `10^(-3) mol% GaCl_(3)`, what is the concentration of the cation vacancies?

Text Solution

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`100 mol` of `NaCl` are doped with `10^(-3) mol` of `GaCl_(3)`.
`:. 1 mol` of `NaCl` is doped with `GaCl_(3) = (10^(-3))/(100) = 10^(-5) mol`
As one `Ga^(3+)` ion is introduced, three `Na^(o+)` have to be removed to maintain the electrical neutrality. So as one vacancyis filled by `Ga^(3+)`, two cation vacancies are formed.
`:.` Concentration of cation vacancy
`= 2 xx 10^(-5)` mol/mol of `NaCl`
`= 2 xx 10^(-5) xx 6.023 xx 10^(23) mol^(-10)`
`= 12.046 xx 10^(-18) mol^(-1)`
`= 1.2046 xx 10^(-19) mol^(-1)`
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