If `NaCI` is droped with `10^(-3)` mole of `SrCI_(2)` then number of cationic vacancies is

Options

A. `6.02 xx 10^(-18) mol^(-1)`
B. `10^(-5) mol^(-1)`
C. `6.02 xx 10^(20) mol^(-1)`
D. `6.02 xx 10^(18) mol^(-1)`

Answer

D

Explanation

Due to the addition of `SeCI_(2) ` each `Sr^(2+)` ion replace two `Na^(+)` ions but occupies one `Na^(+)` lattice point .Thus , this exchanged of `Na^(+)` ion by `Sr^(2)` ion makes one catiomic vacancy `SeCI_(2)` doped `= 10^(-3)` and per `100 mol`
` = 10^(-3) mol per 1 mol`
Cation vacancies `= 10^(-3) xx 6.02 xx 10^(23)`
Total `= 6.02 xx 10^(14)` cationic vacancies `mol^(-1)`