If NaCl is doped with `10^(-4) mol% "of" SrCl_(2)`the concentration of cation vacancies will be
`(N_(A)=6.02times10^(23)mol^(-1))`

Options

A. `6.02times10^(23) mol^(-1)`
B. `6.02times10^(15) mol^(-1)`
C. `6.02times10^(16) mol^(-1)`
D. `6.02times10^(17) mol^(-1)`

Answer

D

Explanation

For each `Sr^(2+)`ion introduced, one vacancy is created because `2Na^(+)`ions are removed and one of the two vacant sites is occupied by `Sr^(2+)`. Doping with `10^(-4)` mol % `SrCl_(2)` means 100 moles of NaCl are doped with `10^(-4)` mole of `SrCl_(2)`.
`therefore SrCl_(2)" droped per mole of " NaCl = 10^(-4)//100`
=`10^(-6) "mole" = 10^(-6)times(6.02times10^(23))Sr^(2+) Sr^(2+)ions`
`therefore"concentration of cation vacancies"=6.02times10^(17) mol^(-1)`