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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) = 6.022 xx 10^(23) mol^(-1))`

A

`6.02 xx 10^(14) mol^(-1)`

B

`6.02 xx 10^(15) mol^(-1)`

C

`6.02 xx 10^17 mol^(-1)`

D

`6.02 xx 10^16 mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
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