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if AgCl is dopped with 0.001 mol percent...

if AgCl is dopped with 0.001 mol percent of `CdCl_(2)` the number of cation vacancies per mol of Agcl will be

A

`6.02xx10^(23)`

B

`6.02xx10^(20)`

C

`6.02xx10^(18)`

D

`3.01xx10^(18)`

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