1.

I mol of NaCl is doped with 10^(-5) mole of SrCl_(2) .The number of cationic vacancies in the crystal lattice will be

Answer»

`6.022 xx 10^(18)`
`6.022 xx 10^(15)`
`6.022 xx 10^(23)`
`12.044 xx 10^(20)`

SOLUTION :(a): 1 MOLE of NaCl is doped with `10^(-5)` mole of `SrCl_(2)`.
As each `Sr^(2+)` ion introduces one cationic vacancy, therefore, number of cationic VACANCIES
`= 10^(-5) xx 6.02 xx 10^(23) mol^(-1) = 6.02 xx 10^(18) mol^(-1)`


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