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If NaCl is doped with 10^(-3) mol % of S...

If NaCl is doped with `10^(-3)` mol `%` of `SrCl_(2)`, what is the concentration of cation vacancies?

A

`6.022xx10^(16)" mol"^(-1)`

B

`6.022 xx 10^(17)"mol"^(-1)`

C

`6.022xx10^(13)"mol"^(-1)`

D

`6.022xx10^(15)"mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B

Doping by each `Sr^(2+)` ion will create one cation vacancy
:. Concentration of cation vacancies
`=6.022xx10^(23)xx10^(-14)/100`
`=6.022xx10^(17)`
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