Question

Calculate the number of sodium atoms present in $5.3\text{g}$ of $N{a}_{2}C{O}_3$.

• A. $12.17\times {10}^{23}\text{atoms}$
• B. $6.023\times {10}^{23}\text{atoms}$
• C. $6.023\times {10}^{22}\text{atoms}$
• D. $12.17\times {10}^{22}\text{atoms}$

Answer 1

The correct option is C $6.023\times {10}^{22}\text{atoms}$

Gram molar mass of $N{a}_{2}C{O}_3=2\times 23+12+3\times 16=106\text{g}$
$106\text{g}$ of $N{a}_{2}C{O}_3$ = 1 mol
$5.3\text{g}$ of $N{a}_{2}C{O}_3$ = $\frac{1\times 5.3}{106}=0.05\text{mol}$
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
$=6.023\times {10}^{22}$ $Na$ atoms