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. `6.02 xx 10^15 "mol"^(-1)`
B. `6.02 xx 10^15 "mol"^(-1)`
C. `6.02 xx 10^17 "mol"^(-1)`
D. `6.02 xx 10^14 "mol"^(-1)`

Answer

C

Explanation

Number of cationic vacancies = charge difference x number of impurity ions
` = (1 xx 10^(-4) xx 6.02 xx 10^23)/(100) = 6.02 xx 10^17`