If `NaCI` is droped with `10^(-4) "mole"% of SrCI_(2)` then number of cationic vacancies will be

Options

A. `6.02 xx 10^(16) mol^(-1)`
B. `6.02 xx 10^(17) mol^(-1)`
C. `6.02 xx 10^(14) mol^(-1)`
D. `6.02 xx 10^(15) mol^(-1)`

Answer

b

Explanation

What `SrCI_(2)` doped with `NaCI`
One `Sr^(2+)` replaces two `Na^(+)` ions and occupies a lattice point and produces one cation vacancy
`100 mol NaCI `will have `10^(-4)` cation vacancy
` 1 mol = (10^(4))/(100) = 10^(6) mol`
Number of cation vacancies `= 10^(-6) xx 6.02 xx 10^(23)`
`= 6.02 xx 10^(17)` atoms