Calculate the number of sodium atoms present in $5.3 g$ of $N a_{2} C O_{3}$ .

12.17 \times 10^{23} atoms

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6.023 \times 10^{23} atoms

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6.023 \times 10^{22} atoms

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12.17 \times 10^{22} atoms

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Solution

The correct option is C $6.023 \times 10^{22} atoms$
Gram molar mass of $N a_{2} C O_{3} = 2 \times 23 + 12 + 3 \times 16 = 106 g$
$106 g$ of $N a_{2} C O_{3}$ = 1 mol
$5.3 g$ of $N a_{2} C O_{3}$ = $\frac{1 \times 5.3}{106} = 0.05 mol$
Now,1 mol of $N a_{2} C O_{3}$ contains = $2 \times 6.023 \times 10^{23}$ sodium atoms
Therefore, 0.05 mol of $N a_{2} C O_{3}$ contains = $0.05 \times 2 \times 6.023 \times 10^{23}$ sodium atoms
$= 6.023 \times 10^{22}$ $N a$ atoms

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