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If NaCl crystals are doped with 2 xx 10^(-3) mol per cent of SrCl_(2), calculate the cation vacancies per mole. |
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Answer» Solution :Doping of `NaCl` with `2xx10^(-3)` mol per cent of `SrCl_(2)` means 100 MOLES of `NaCl` are dropped with `2xx10^(-3)` moles of `SrCl_(2)` or 1 mole of `NaCl` is doped with `=2xx10^(-5)` mole of `SrCl_(2)`. Each `Sr^(2+)` will occupy the place of `Na^(+)` and displace one `Na^(+)` from crystal lattice to create cation vacancies. Cation vacancies = Number of `Sr^(2+)` ion added `=2xx10^(-5)mol=2xx10^(-5)xx6.023xx10^(23)=12.046xx10^(18)=1.2046xx10^(19)mol^(-1)`. |
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