Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

$25.3\text{ g}$ of Sodium carbonate $\text{Na}_2\text{CO}_3$ is dissolved in enough water to make $250\text{ mL}$ of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, $\text{Na}^+$ and carbonate ion $\text{CO}_3^{2-}$ are respectively (Molar mass of $\text{Na}_2\text{CO}_3 = 106\text{ g mol}^{-1}$ )

$0.955\text{ M}$ and $1.910\text{ M}$

$1.910\text{ M}$ and $0.955\text{ M}$

$1.90\text{ M}$ and $1.910\text{ M}$

$0.477\text{ M}$ and $0.477\text{ M}$

Step-by-Step Solution

Moles of $\text{Na}_2\text{CO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{25.3\text{ g}}{106\text{ g mol}^{-1}} = 0.2387\text{ mol}$ . Volume of solution $= 250\text{ mL} = 0.250\text{ L}$ . Molarity of $\text{Na}_2\text{CO}_3 = \frac{0.2387\text{ mol}}{0.250\text{ L}} = 0.9548\text{ M} \approx 0.955\text{ M}$ . The dissociation reaction of sodium carbonate is: $\text{Na}_2\text{CO}_3 \rightarrow 2\text{Na}^+ + \text{CO}_3^{2-}$ Since $1\text{ mole}$ of $\text{Na}_2\text{CO}_3$ gives $2\text{ moles}$ of $\text{Na}^+$ and $1\text{ mole}$ of $\text{CO}_3^{2-}$ , Concentration of $\text{Na}^+ = 2 \times 0.955\text{ M} = 1.910\text{ M}$ . Concentration of $\text{CO}_3^{2-} = 1 \times 0.955\text{ M} = 0.955\text{ M}$ .

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