Question

Calculate the number of sodium and carbon atoms in 8 grams of sodium carbonate $\left(N{a}_{2}C{O}_3\right)$.

• A. Sodium atoms - $0.9\times {10}^{23}$ Carbon atoms - $0.45\times {10}^{23}$
• B. Sodium atoms - $20\times {10}^{23}$ Carbon atoms - $5\times {10}^{23}$
• C. Sodium atoms - $1\times 6\times {10}^{23}$ Carbon atoms - $5\times {10}^{23}$
• D. Sodium atoms - $4\times {10}^{23}$ Carbon atoms - $5\times {10}^{23}$

Answer 1

The correct option is A

Sodium atoms - $0.9\times {10}^{23}$

Carbon atoms - $0.45\times {10}^{23}$

Molecular mass of $N{a}_{2}C{O}_3$

= 2(Na) + C + 3(0)

= 2(23) + 12 + 3(16) = 106 gms

In 106 gm number of sodium atoms $=6.023\times {10}^{23}\times 2$

In 5.3 g number of sodium atoms $=\frac{6.02\times {10}^{23}\times 2}{106}\times$
$=0.9\times {10}^{23}$

Number of carbon atoms in 8 gms

$=\frac{6.023\times {10}^{23}}{106}\times 8=0.45\times {10}^{23}$