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^(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

One cation of `Sr^(2+)` would create one cation vacancy in NaCl. Therefore , the number of cation vacancies created in the lattice of NaCl is equal to the number of divalent `Sr^(2+)` ions added.
No. of moles of cationic vacancies
`= (10^(-4))/(10^2) = 10^(-6) mol`
No. of cation vacancies
`=10^(-6) xx 6.02 xx 10^(23)`
`=6.02 xx 10^(17)`.