$25.3 \mathrm{~g}$ of sodium carbonate, $\mathrm{Na}_2 \mathrm{CO}_3$ is dissolved in enough water to make…
$25.3 \mathrm{~g}$ of sodium carbonate, $\mathrm{Na}_2 \mathrm{CO}_3$ is dissolved in enough water to make $250 \mathrm{~mL}$ of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, $\mathrm{Na}^{+}$and carbonate ion, $\mathrm{CO}_3^{2-}$ are respectively (Molar mass of $\mathrm{Na}_2 \mathrm{CO}_3=106 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$0.955 \mathrm{M}$ and $1.910 \mathrm{M}$
$1.910 \mathrm{M}$ and $0.955 \mathrm{M}$
$1.90 \mathrm{M}$ and $1.910 \mathrm{M}$
$0.477 \mathrm{M}$ and $0.477 \mathrm{M}$
Solution
$\begin{aligned}
\text { Molarity } & =\frac{\text { number of moles of solute }}{\text { volume of solution (in mL) }} \times 1000 \\
& =\frac{25.3 \times 1000}{106 \times 250}=0.9547 \approx 0.955 \mathrm{M}
\end{aligned}$
$\mathrm{Na}_2 \mathrm{CO}_3$ in aqueous solution remains dissociated as
Since, the molarity of $\mathrm{Na}_2 \mathrm{CO}_3$ is $0.955 \mathrm{M}$, the molarity of $\mathrm{CO}_3^{2-}$ is also $0.955 \mathrm{M}$ and that of $\mathrm{Na}^{+}$is $2 \times 0.955=1.910 \mathrm{M}$
Since, the molarity of $\mathrm{Na}_2 \mathrm{CO}_3$ is $0.955 \mathrm{M}$, the molarity of $\mathrm{CO}_3^{2-}$ is also $0.955 \mathrm{M}$ and that of $\mathrm{Na}^{+}$is $2 \times 0.955=1.910 \mathrm{M}$