If `NaCl` is doped with `10^(-4)mol%`of `SrCl_(2)` the concentration of cation vacancies will be `(N_(A)=6.02xx10^(23)mol^(-1))`

Options

A. `6.022xx10^(16)" mol"^(-1)`
B. `6.022 xx 10^(17)"mol"^(-1)`
C. `6.022xx10^(14)"mol"^(-1)`
D. `6.022xx10^(15)"mol"^(-1)`

Answer

B

Explanation

Doping by each `Sr^(2+)` ion will create one cation vacancy
:. Concentration of cation vacancies
`=6.022xx10^(23)xx10^(-14)/100`
`=6.022xx10^(17)`