In NaCl is doped with ` 10^(-4)` mol % of ` ScCl_(2)` , the concentration of cation vacancies will be ` (N_(A) = 6.02 xx 10^(23) mol^(-1))`

Options

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

Answer

b

Explanation

For each ` Sr^(2+)` ion introduced, one catio 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)`
`SrCl_(2)` doped per mole of NaCl = `10^(-4) //100`
` 10^(-6)` mole
=`10^(-6) xx ( 6.02 xx 10^(23)) Sr^(2+)` ions
` = 6.02 xx 10^(17) Sr^(2+)` ions
Hence, concetration of cation vacancies
` 6.02 xx 10^(17) "mol"^(-1)`