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If NaCl is doped with 10^(-3) mol % of SrCl_(2) what is the concentration of cation vacancy? |
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Answer» Solution :ONE cation of `SR^(2+)` would create one cation vacancy in NaCl. `:.`Concentration of cation vacancy on being doped with `10^(-3)` mol% `SrCl_(2)` `= 10 ^(-3) mol%` `=(10 ^(-3))/(100) "mol"` `=(10 ^(-3))/(100) XX 6.02 xx 10^(23)` cation vacancies PER mole `= 6.02 xx10^(18)`cation vacancies per mole. |
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