The number of sodium atoms in $2$ moles of sodium ferrocyanide, $N a_{4} [F e {(C N)}_{6}]$ is:

12 \times 10^{23}

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26 \times 10^{23}

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34 \times 10^{23}

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48 \times 10^{23}

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Solution

The correct option is D $48 \times 10^{23}$
The number of sodium atoms in $1$ molecule of sodium ferrocyanide, $N a_{4} [F e {(C N)}_{6}]$ is 4.
The number of sodium atoms in $1$ moles of sodium ferrocyanide, $N a_{4} [F e {(C N)}_{6}]$ is 4 times avogadros number. It is equal to $4 \times 6.0 \times 10^{23}$ .
The number of sodium atoms in $2$ moles of sodium ferrocyanide, $N a_{4} [F e {(C N)}_{6}]$ is 8 times avogadros number. It is equal to $8 \times 6.0 \times 10^{23}$ or $48 \times 10^{23}$ .
Thus, option $D$ is correct.

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