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

Answer

1

Explanation

If `NaCl` is doped with `10^(-4)` mol% of `SrCl_(2)` it implies that for `100` moles of `NaCl` we add `10^(-4)` moles of `SrCl_(2)` thus `1` mole of `NaCl` is doped with `10^(-4)//100` mols of `SrCl_(2)`.
Total number of `SrCl_(2)` formula units
`=(6.02xx10^(23)"formula unit" mol^(-1))(10^(-6)mol)`
Total number of `Sr^(2+)` cations
`=(6.02xx10^(23)"formula units" mol^(-1))(10^(-6)mol)`
`(1 Sr^(2+)//"formula unit")`
Since every `Sr^(2+)` cation introudces one cation vacancy the concentration fo cation vacncies will be
`=(6.02xx10^(23)("formula units")/("mol"))(10^(-6)mol)((1Sr^(2+))/("formula unit"))`
`= ((1"cation vacancy")/(Sr^(2+)ion))`
`= 6.02xx10^(17)mol^(-1)`