Question

21.2 g of sodium carbonate $\left(N{a}_{2}C{O}_3\right)$ is dissolved in enough water to make 250 mL of a solution. If it dissociates completely, the molar concentration of sodium ions $\left(N{a}^{+}\right)$ and the carbonate ions $\left(C{O}_3^{2-}\right)$ are respectively:
(Molar mass of $N{a}_{2}C{O}_3$ = 106 $gmo{l}^{-1}$)

• A. 0.8 M and 1.6 M
• B. 1.6 M and 0.8 M
• C. 1.6 M and 1.6 M
• D. 0.8 M and 0.8 M

Answer 1

The correct option is B 1.6 M and 0.8 M

Number of moles of $N{a}_{2}C{O}_3=\frac{\text{given mass (g)}}{\text{molar mass (g/mol)}}=\frac{21.2}{106}=0.2mol$

$\text{Molarity}=\frac{\text{moles of}N{a}_{2}C{O}_3}{\text{Total volume of solution in mL}}\times 1000\text{Molarity}=\frac{0.2}{250}\times 1000=0.8M$

According to the reaction: $N{a}_{2}C{O}_3\to 2N{a}^{+}+C{O}_3^{2-}$
$1M$ $N{a}_{2}C{O}_3$ will dissociate into $2M$ of $N{a}^{+}$ and $1M$ of $C{O}_3^{2-}$
0.8 M of $N{a}_{2}C{O}_3$ dissociates into $2\times 0.8$ M $N{a}^{+}=1.6MN{a}^{+}$
$0.8$ M $N{a}_{2}C{O}_3$ dissociates to $0.8$ M of $C{O}_3^{2-}$ .