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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

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

B

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

C

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

D

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

Text Solution

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The correct Answer is:
C
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