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. `6.02xx10^14 "mol"^(-1)`
B. `6.02xx10^15 "mol"^(-1)`
C. `6.02xx10^16 "mol"^(-1)`
D. `6.02xx10^17 "mol"^(-1)`

Answer

D

Explanation

For each `Sr^(2+)` ion introduced, one cation vacancy is created because `2 Na^+` ions are removed and one vacant site is occupied by `Sr^(2+)` . Doping with `10^(-4)` mol % of `SrCl_2` means 100 moles of NaCl are doped with `10^(-4)` mole of `SrCl_2`.
`therefore SrCl_2` doped per mole of NaCl=`10^(-4)//100`
=`10^(-6)` mole
=`10^(-6)xx (6.02xx10^23) Sr^(2+)` ions
`=6.02xx10^17 Sr^(2+)` ions
Hence, concentration of cation vacancies =`6.02xx10^17 mol^(-1)`