If NaCl is doped with `10^(-4)` mol % of `SrCl_(2)` , the concentration of cation vacancies will be `( N_(A) = 6.022 xx 10^(23) mol^(-1))`

Options

A. `6.02 xx 10^(18) mol^(-1)`
B. `6.02 xx 10^(17) mol^(-1)`
C. `6.02 xx 10^(14) mol^(-1)`
D. `6.02 xx 10^(15) mol^(-1)`

Answer

B

Explanation

One cation of `Sr^(2+)` 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 divalent `Sr^(2+)` ions added.
No. of moles of cationic vacancies ` = ( 10^(-4))/( 10^(2)) = 10^(-6)` mol
No. of cation vacancies
` = 10^(-6) xx 6.02 xx 10^(23) = 6.02 xx 10^(17)`