Answer
$4.85\times 10^{24}atoms$
Work Step by Step
We can solve it as
$123g Mg(CN)_2\times \frac{1 mol\space Mg(CN)_2}{76.3g\space Mg(CN)_2}\times \frac{6.023\times 10^{23}molecules }{1 \space mol \space Mg(CN)_2}\times\frac{5atoms}{1 molecule}=4.85\times 10^{24}atoms$
