If $NaCl$ is doped with ${10}^{-2}mole\%SrC{l}_2,$ then the number of cation vacancies is $6.023\times {10}^x$ then find the $\frac{x}{9}$ value (round the nearest integer value) is____

100 moles of $NaCl$ are doped with 0.01 moles of $SrC{l}_2$

1 mole of $NaCl$ is doped with $=\frac{0.01}{100}={10}^{-4}molesofSrC{l}_2$

$\therefore {10}^{-4}$ moles of $SrC{l}_2\left(or\right){10}^{-4}$ 4moles of $S{r}^{+2}$ cations

${10}^{-4}$ moles of $S{r}^{+2}$ will be create ${10}^{-4}$ moles of cation vacancies

$\therefore$the number of cation vacancies  $={10}^{-4}x6.023x{10}^{23}=6.023x{10}^{19}=\frac{x}{9}=2.1$