`25.3 g` sodium carbonate, `Na_(2)CO_(3)`, was dissolved in enough water to make `250 mL` of solution. If sodium carbonate dissociates completely, molar concentration of `Na^(+)` and carbonate ions are respectively:

Options

A. (a)`0.9555 M and 1.910 M`
B. (b)`1.910 M and 0.955 M`
C. (c )`1.90 M and 1.090 M`
D. (d)`0.477 M and 0.477 M`

Explanation

To find the molar concentration of sodium ions (Na⁺) and carbonate ions (CO₃²⁻) in a solution of sodium carbonate (Na₂CO₃), we will follow these steps:

### Step-by-Step Solution:

1. **Calculate the Molar Mass of Na₂CO₃:**
- The molar mass of sodium (Na) is approximately 23 g/mol.
- The molar mass of carbon (C) is approximately 12 g/mol.
- The molar mass of oxygen (O) is approximately 16 g/mol.
- Therefore, the molar mass of Na₂CO₃ = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 g/mol.

2. **Calculate the Number of Moles of Na₂CO₃:**
- We have 25.3 g of Na₂CO₃.
- Number of moles = mass (g) / molar mass (g/mol).
- Number of moles of Na₂CO₃ = 25.3 g / 106 g/mol = 0.238 moles.

3. **Determine the Moles of Ions Produced:**
- When Na₂CO₃ dissociates, it produces 2 moles of Na⁺ ions and 1 mole of CO₃²⁻ ions.
- Therefore, from 0.238 moles of Na₂CO₃:
- Moles of Na⁺ = 2 × 0.238 = 0.476 moles.
- Moles of CO₃²⁻ = 0.238 moles.

4. **Convert Volume from mL to Liters:**
- The volume of the solution is given as 250 mL.
- To convert to liters, we use the conversion: 1 L = 1000 mL.
- Volume in liters = 250 mL × (1 L / 1000 mL) = 0.250 L.

5. **Calculate the Molar Concentration of Na⁺:**
- Molarity (M) = moles of solute / volume of solution in liters.
- Molar concentration of Na⁺ = 0.476 moles / 0.250 L = 1.904 M (approximately 1.910 M).

6. **Calculate the Molar Concentration of CO₃²⁻:**
- Molar concentration of CO₃²⁻ = 0.238 moles / 0.250 L = 0.952 M (approximately 0.955 M).

### Final Results:
- The molar concentration of Na⁺ is approximately **1.910 M**.
- The molar concentration of CO₃²⁻ is approximately **0.955 M**.