$25.3 \mathrm{~g}$ of sodium carbonate, $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is dissolved in enough water to…
water to make $250 \mathrm{~mL}$ of solution. If sodium carbonate
dissociates completely, molar concentration of sodium ions, $\mathrm{Na}^{+}$ and carbonate ions, $\mathrm{CO}_{3}^{2-}$ respectively are (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
$\mathrm{Na}_{2} \mathrm{CO}_{3}=\frac{25.3}{106} \times \frac{1000}{250}=0.955 \mathrm{M}$
$\left[\mathrm{Na}^{+}ight]=2 \times 0.955=1.91 \mathrm{M}$
$\left[\mathrm{CO}_{3}^{2-}ight]=0.955 \mathrm{M}$
Asked in: JEE-TOPICTESTS-CHEMISTRY