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

If `NaCl` is doped with `10^(-4)mol%`of `SrCl_(2)` the concentration of cation vacancies will be `(N_(A)=6.02xx10^(23)mol^(-1))`

A

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

B

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

C

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

D

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

Text Solution

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The correct Answer is:
B
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