How many sodium ions are present in a aqueous solution of sodium carbonate containing 2.65 g?

here's the solution for the next question.

To find the number of sodium ions, we first need to calculate the moles of sodium carbonate, then consider its dissociation, and finally use Avogadro's number. We will use the following approximate relative atomic masses: Na = 23, C = 12, O = 16. Avogadro's number ($N_A$) = $6.022 \times 10^{23} \text{ mol}^{-1}$.

Step 1: Calculate the molar mass of sodium carbonate (Na₂CO₃). $\text{Molar mass of Na₂CO₃} = (2 \times 23) + (1 \times 12) + (3 \times 16)$ $\text{Molar mass of Na₂CO₃} = 46 + 12 + 48 = 106 \text{ g/mol}$

Step 2: Calculate the number of moles of sodium carbonate in 2.65 g. $\text{Moles of Na₂CO₃} = \frac{\text{given mass}}{\text{molar mass}}$ $\text{Moles of Na₂CO₃} = \frac{2.65 \text{ g}}{106 \text{ g/mol}} = 0.025 \text{ mol}$

Step 3: Determine the number of sodium ions produced per mole of Na₂CO₃. When sodium carbonate dissolves in water, it dissociates as follows: $\text{Na₂CO₃(s)} \longrightarrow \text{2Na⁺(aq)} + \text{CO₃²⁻(aq)}$ From the equation, 1 mole of Na₂CO₃ produces 2 moles of Na⁺ ions.

Step 4: Calculate the number of moles of sodium ions. $\text{Moles of Na⁺ ions} = 2 \times \text{Moles of Na₂CO₃}$ $\text{Moles of Na⁺ ions} = 2 \times 0.025 \text{ mol} = 0.050 \text{ mol}$

Step 5: Calculate the total number of sodium ions using Avogadro's number. $\text{Number of Na⁺ ions} = \text{Moles of Na⁺ ions} \times N_A$ $\text{Number of Na⁺ ions} = 0.050 \text{ mol} \times 6.022 \times 10^{23} \text{ ions/mol}$ $\text{Number of Na⁺ ions} = 3.011 \times 10^{22} \text{ ions}$ The number of sodium ions present is $\boxed{\text{3.011} \times 10^{22}}$.