1.

If NaCl is doped with 10^(-2) mol % SrCl_2, what is the concentration of cation vacancies?

Answer»


SOLUTION :Doping of NaCl with `10^(-2)` mole % of `SrCl_2` means that 100 MOLES of NaCl are doped with 10 mole of `SrCl_2`.
`:.` 1 mole of NaCl is doped with `SrCl_2=(10^(-2))/(100)= 10^(-4)` MOL
As each `Sr^(2+)` ion introduces one cation vacancy, therefore, concentration of cation vacancies = `10^(-4) mol//mol` of NaCl = `10^(-4) xx 6.02 xx 10^(23)` cation vacancies `mol^(-1) = 6.02 xx 10^(19)` cation vacancies `mol^(-1)`.


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