The number of sodium atoms in 2 moles of sodium ferrocyanide Na-4-Fe-CN-6- is-2- 6-023- 10-236-023- 10-238- 6-023- 10-234- 6-023- 10-23

The correct option is C 8× 6.023× 10^23Since 1 formula unit of nbsp;Na_4[Fe(CN)_6] contains 4 Na atoms, 1 mole of the same will contain 4 moles of sodium ato ...

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Standard XII

Chemistry

Ammonia

The number of...

Question

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

2 \times 6.023 \times 10^{23}

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6.023 \times 10^{23}

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8 \times 6.023 \times 10^{23}

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4 \times 6.023 \times 10^{23}

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Solution

The correct option is C $8 \times 6.023 \times 10^{23}$
Since $1$ formula unit of $N a_{4} [F e (C N)_{6}]$ contains 4 $N a$ atoms, $1$ mole of the same will contain $4$ moles of sodium atoms.
Similarly, $2$ mol of compound will contain $8$ mol $N a$ atoms
$= 8 \times 6.023 \times 10^{23}$ atoms.

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The number of sodium atoms in 2 moles of sodium ferrocyanide Na4[Fe(CN)6] is:

The correct option is C 8×6.023×1023Since 1 formula unit of Na4[Fe(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 =8×6.023×1023 atoms.

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

The correct option is C $8 \times 6.023 \times 10^{23}$
Since $1$ formula unit of $N a_{4} [F e (C N)_{6}]$ contains 4 $N a$ atoms, $1$ mole of the same will contain $4$ moles of sodium atoms.
Similarly, $2$ mol of compound will contain $8$ mol $N a$ atoms
$= 8 \times 6.023 \times 10^{23}$ atoms.