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. `6.02 xx 10^(15) "mol"^(-1)`
B. `6.02 xx 10^(16) "mol"^(-1)`
C. `6.02 xx 10^(17) "mol"^(-1)`
D. `6.02 xx 10^(14) "mol"^(-1)`

Answer

C

Explanation

One cation of `Sr^(++)` would create one cation vacancy in NaCl therefore the number of cation vacancies created in the lattice of NaCl is equal to the number of `Sr^(++)` ions added. Concentration of cation vacancy on being doped with
`10^(-4) "mole" % SrCl_(2) = 10^(-4)/100 = 10^(-6)` mol
No. of `Sr^(++)` ions, in `10^(-6) xx 6.022 xx 10^(23)`
`=6.022 xx 10^(17) Sr^(++)` ions.
No. of cation vacancies `= 6.022 xx 10^(17)`