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प्रश्न
If NaCl is doped with 10−3 mol % of SrCl2, what is the concentration of cation vacancies?
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उत्तर
When NaCl is doped with SrCl2, each additional Sr2+ ion eliminates two Na+ ions, creating two vacancies, one occupied by Sr2+ and the other unoccupied. Thus, adding one Sr2+ ion results in one cation vacancy.
Doping of NaCl with 1 × 10−3 rnol % of SrCl2 means that 1 × 10−3 moles of NaCl are added to 100 moles of NaCl.
∴ 1 mole of NaCl contains SrCl2 = `(1 xx 10^-3)/100`
= 1 × 10−5 moles
= 1 × 10−5 × 6.022 × 1023 molecules
= 6.022 × 1018 molecules
∴ No. of Sr2+ ions present in one mole of NaCl = No. of cation vacancies in one mole of NaCl
= 6.022 × 1018
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