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If NaCl is doped with 10^(-2) mol percentage of strontium chloride, what is the concentration of cation vacancy ? |
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Answer» Solution :We KNOW that two `Na^(+)`ions are replaced by each of the `Sr^(2+)` ions while `SrCl_(2)` is doped with NaCl. But in this case, only one lattice POINT is occupied by each of the `Sr^(2+)` ions and PRODUCE one cation vacancy. Here `10^(-2)` mole of `SrCI_(2)` is doped with 100 moles of NaCl. Thus, cation vacancies produced by NaCl = `10^(-2)` mol. Since, 100 moles of NaCl produces cation vacancies after doping = `10^(-2)` mol. Therefore, 1 mole of NaCl will produce cation vacancies after doping `=(10^(-2))/(100)=10^(-4)"mol"` `therefore"Total cationic vacanices,"` `=10^(-4)xx"Avogadro.s NUMBER"` `=10^(-4)xx6.023xx10^(23)` `=6.023xx10^(19)" vacancies"` |
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