The number of sodium atoms in 2 mol of sodium ferrocyanide is

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The number of sodium atoms in 2 mol of sodium ferrocyanide is

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$26 \times 10^{23}$

$34 \times 10^{23}$

$48 \times 10^{23}$

$12 \times 10^{23}$

answer is A.

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Detailed Solution

As we know that four sodium atoms are present in sodium ferrocyanide ${Na}_{4} [Fe (CN)_{6}]$

Hence, number of Na atoms - No. of moles x number of atom

× Avogadro's number

$2 \times 4 \times 6.023 \times 10^{23} = 48 \times 10^{23}$

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