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If NaCl is doped with 10^(4) mol % of Sr...

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

A

`6.02 xx10^(14)mol^(-1)`

B

`6.02 xx10^(15)mol^(-1)`

C

`6.02 xx10^(16)mol^(-1)`

D

`6.02 xx10^(17)mol^(-1)`

Text Solution

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The correct Answer is:
D
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