The number of sodium atoms in 2 moles of sodium ferrocyanide $\left(N{a}_4\right[Fe\left(CN{)}_6\right])$ is:
(Consider Avogadro's number = $6\times {10}^{23})$

The correct option is A $48\times {10}^{23}$
We know that four sodium atoms are present in sodium ferrocyanide $N{a}_4\left[Fe\right(CN{)}_6]$.
Hence, the total number of $Na$ atoms in 2 moles of $N{a}_4\left[Fe\right(CN{)}_6]$
= number of moles $\times$ number of atom $\times$ Avogadro's number
$=2\times 4\times 6\times {10}^{23}=48\times {10}^{23}$