Question

The number of sodium atoms in $2$ moles of sodium ferrocyanide $N{a}_4\left[Fe\right(CN{)}_6]$ is:

• A. $2\times 6.023\times {10}^{23}$
• B. $6.023\times {10}^{23}$
• C. $8\times 6.023\times {10}^{23}$
• D. $4\times 6.023\times {10}^{23}$

Answer 1

The correct option is C $8\times 6.023\times {10}^{23}$

Since $1$ formula unit of $N{a}_4\left[Fe\right(CN{)}_6]$ contains 4 $Na$ atoms, $1$ mole of the same will contain $4$ moles of sodium atoms.
Similarly, $2$ mol of compound will contain $8$ mol $Na$ atoms
$=8\times 6.023\times {10}^{23}$ atoms.