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If NaCl is doped with 10^(-4)"mol"% of s...

If `NaCl` is doped with `10^(-4)"mol"%` of `srCl_(2)`, the concentration of cation vacancies will be `(N_(A)=6.023xx10^(23)"mol"^(-1))`

A

`6.02times10^(23) mol^(-1)`

B

`6.02times10^(15) mol^(-1)`

C

`6.02times10^(16) mol^(-1)`

D

`6.02times10^(17) mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
For each `Sr^(2+)`ion introduced, one vacancy is created because `2Na^(+)`ions are removed and one of the two vacant sites is occupied by `Sr^(2+)`. Doping with `10^(-4)` mol % `SrCl_(2)` means 100 moles of NaCl are doped with `10^(-4)` mole of `SrCl_(2)`.
`therefore SrCl_(2)" droped per mole of " NaCl = 10^(-4)//100`
=`10^(-6) "mole" = 10^(-6)times(6.02times10^(23))Sr^(2+) Sr^(2+)ions`
`therefore"concentration of cation vacancies"=6.02times10^(17) mol^(-1)`
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