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If NaCl is doped with 10^(-3) mole percent SrCl_(2), what will be the concentration of cation vacancies ? (N_(A) = 6.02 xx 10^(23) mol^(-1)) |
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Answer» Solution :Every `Sr^(2+)` ion causes ONE cation vacancy (because two `Na^(+)` ions are REPLACED by one `Sr^(2+)`) Therefore, introduction of `10^(-3)` moles of `SrCl_(2)` per 100 moles of NaCl would introduce `10^(-3)` mole cation VACANCIES in 100 moles of NaCl. No. of vacancies per mole of NaCl `= (10^(-3))/(10000) xx 6.02 xx 10^(23) = 6.02 xx 10^(18)` vacancies. |
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