Question

Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.

Answer 1

$$\begin{gathered} Noofmole=\frac{Givenmassing}{Molarmass} \\ Noofmole=\frac{5.3}{106}=0.05 \\ Noofmolecules=mole\times avogadronumber \\ =0.05\times 6.022\times {10}^{23} \\ =0.3011\times {10}^{23} \\ NoofatomsofSodiuminN{a}_{2}C{O}_3=0.3011\times {10}^{23}\times 2=0.6022\times {10}^{23} \\ NoofatomsofCarboninN{a}_{2}C{O}_3=0.3011\times {10}^{23}\times 1=0.3011\times {10}^{23} \\ NoofatomsofoxygeninN{a}_{2}C{O}_3=0.3011\times {10}^{23}\times 3=0.9033\times {10}^{23} \end{gathered}$$