The exact number of sodium ions present in 53 g of anhydrous sodium carbonate ($\mathrm{Na}_2\mathrm

Question

The exact number of sodium ions present in 53 g of anhydrous sodium carbonate ($\mathrm{Na}_2\mathrm{CO}_3$) is ($N_A$ represents Avogadro's number, Molar mass of $\mathrm{Na}_2\mathrm{CO}_3 = 106 \mathrm{~g/mol}$)

Professor PC Thomas Classes · Answer

(a) : The number of moles of $\mathrm{Na}_2\mathrm{CO}_3$ is calculated as $	ext{Mass} / 	ext{Molar mass} = 53 / 106 = 0.5 \mathrm{~mol}$. The chemical formula indicates that one formula unit of $\mathrm{Na}_2\mathrm{CO}_3$ completely dissociates to yield $2 \mathrm{Na}^+$ ions. Therefore, the total moles of $\mathrm{Na}^+$ ions is $0.5 	imes 2 = 1.0 \mathrm{~mol}$. This corresponds exactly to $1.0 	imes N_A$ individual sodium ions.