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. `6.023xx10^(15)"mol"^(-1)`
B. `6.023xx10^(16)"mol"^(-1)`
C. `6.023xx10^(17)"mol"^(-1)`
D. `6.023xx10^(14)"mol"^(-1)`

Answer

C

Explanation

Doping of `NaCl` with `10^(-4)"mol"%` of `SrCl_(2)` means,
100 moles of NaCl are doped with `10^(-4)` mol of `SrCl_(2)`.
`therefore` 1 mol of NaCl is doped with ltBrgt `SrCl_(2)=(10^(-4))/(100)=10^(-6)"mole"`
As each `Sr^(2+)` ion introduces one cation vacancy.
`therefore` Concentration of cation vacancies
`10^(-6)"mol/mol of "NaCl`
`=10^(-6)xx6.023xx10^(23)"mol"^(-1)`
`=6.023xx10^(17)"mol"^(-1)`