If $NaCl$ is doped with $10^{-4}$ mole % of $SrCl_2$, the concentrate of cation vacancies will be : $(N_A = 6.02 \times 10^{23}) mol^{-1}$

Question

( A ) $6.02 \times 10^{14} mol^{-1}$
( B ) $6.02 \times 10^{16} mol^{-1}$
( C ) $6.02 \times 10^{15} mol^{-1}$
( D ) $6.02 \times 10^{17} mol^{-1}$