I mol of NaCl is doped with `10^(-5)` mole of `SrCl_(2)` .The number of cationic vacancies in the crystal lattice will be

Options

A. `6.022 xx 10^(18)`
B. `6.022 xx 10^(15)`
C. `6.022 xx 10^(23)`
D. `12.044 xx 10^(20)`

Answer

A

Explanation

(a): 1 mole of NaCl is doped with `10^(-5)` mole of `SrCl_(2)`.
As each `Sr^(2+)` ion introduces one cationic vacancy, therefore, number of cationic vacancies
`= 10^(-5) xx 6.02 xx 10^(23) mol^(-1) = 6.02 xx 10^(18) mol^(-1)`