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

`2Na^(+)` ions are removed and one `Sr^(+ +)` ions is introduced, one cation vancancy is created.
Doping with `10^(-4)` mole % of `SrCl_(2)` means 100 moles of NaCl are doped with `10^(-4)` moles of `SrCl_(4)`.
`(10^(4-))/(100)=10^(-6) "mol"`
Number of `Sr^(++)` ions `=6.023xx10^(23) xx10^(-6)`
`=6.023 xx10^(17) "mol"^(-1)`
Hence number of cation vancaies `=6.023xx10^(17)"mol"^(-1)`