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If NaCl is doped with 10^(-3) mol % of SrCl_(2), what is the concentration of cation vacancies ? |
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Answer» Solution :Let us take one MOLE of `NaCl`. For every `Sr^(2+)` ion INTRODUCED, one `Na^(+)` ion will be removed and one vacancy will be created. Amount of `SrCl_(2)=(10^(-3))/(100)" moles"` `=10^(-5)" moles "=10^(-5)xx6.02xx10^(23)" molecules "=6.02xx10^(18)" molecules"` As each `Sr^(2+)` ion intrduced creates one cation vacancy, CONCENTRATION of cation vacancies `=6.02xx10^(18)` PER mole. |
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