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If NaCl is doped with 10^(-3) mol % SrCl_2, what is the concentration of cation vacancies? |
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Answer» Solution :Doping of NaCl with `10^(-3) mol % SrCl_2`, means that 100 moles of NaCl are doped with `10^(-3)` mol of `SrCl_2`. `:.` 1 mole of NaCl is doped with `SrCl_2 =(10^(-3))/(100) = 10^(-5)` mol As each `SR^(2+)` ION introduces one cation vacancy, THEREFORE concentration of cation vacancies = `10^(-5) mol//mol` of NaCl = `10^(-5) xx 6.02 xx 10^(23)` cation vacancies `mol^(-1)` = `6.02 xx 10^(18)` cation vacancies `mol^(-1)` |
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