If $10^{-4} \, \text{mole} \%$ of $SrCl_2$ is added to $NaCl$,the concentration of cation vacancies created is...... $(N_A = 6.02 \times 10^{23} \, \text{mol}^{-1})$

  • A
    $6.02 \times 10^{15} \, \text{mol}^{-1}$
  • B
    $6.02 \times 10^{16} \, \text{mol}^{-1}$
  • C
    $6.02 \times 10^{17} \, \text{mol}^{-1}$
  • D
    $6.02 \times 10^{14} \, \text{mol}^{-1}$

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