$25.3 \mathrm{~g}$ 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 $\mathrm{Na}_2 \mathrm{CO}_3=106 \mathrm{~g} \mathrm{~mol}^{-1}$ )

• A

  $0.955 \mathrm{M}$ and $1.910 \mathrm{M}$
• Correct answer
  B

  $1.910 \mathrm{M}$ and $0.955 \mathrm{M}$
• C

  $1.90 \mathrm{M}$ and $1.910 \mathrm{M}$
• D

  $0.477 \mathrm{M}$ 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 $2 \times 0.955=1.910 \mathrm{M}$

Hence, the correct answer is option: 'B'.