The number of sodium atoms in $2 \ mol$ of sodium ferrocyanide $Na_4[Fe(CN)_6]$ is .......... $\times 10^{23}$. (Given $N_A = 6.022 \times 10^{23} \ mol^{-1}$)
- A$12$
- B$26$
- C$34$
- D$48$
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