`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. 0.955 M and 1.910 M
B. 1.910 M and 0.955 M
C. 1.90 M and 1.910 M
D. 0.477 M and 0.477 M

Explanation

To solve the problem, we need to calculate the molar concentration of sodium ions (Na⁺) and carbonate ions (CO₃²⁻) in a solution prepared by dissolving 25.3 g of sodium carbonate (Na₂CO₃) in 250 mL of water.

### Step 1: Calculate the molar mass of sodium carbonate (Na₂CO₃)

The molar mass of sodium carbonate can be calculated as follows:
- Sodium (Na): 22.99 g/mol (2 sodium atoms)
- Carbon (C): 12.01 g/mol (1 carbon atom)
- Oxygen (O): 16.00 g/mol (3 oxygen atoms)

Molar mass of Na₂CO₃ = (2 × 22.99) + (1 × 12.01) + (3 × 16.00)
= 45.98 + 12.01 + 48.00
= 105.99 g/mol

### Step 2: Calculate the number of moles of sodium carbonate

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
We can substitute the values:
\[ \text{Number of moles of Na₂CO₃} = \frac{25.3 \text{ g}}{105.99 \text{ g/mol}} \approx 0.238 \text{ moles} \]

### Step 3: Calculate the molarity of sodium carbonate solution

Molarity (M) is defined as the number of moles of solute per liter of solution:
\[ \text{Molarity} = \frac{\text{number of moles}}{\text{volume of solution (L)}} \]
The volume of the solution is 250 mL, which is 0.250 L. Therefore:
\[ \text{Molarity of Na₂CO₃} = \frac{0.238 \text{ moles}}{0.250 \text{ L}} \approx 0.952 \text{ M} \]

### Step 4: Determine the molar concentration of sodium ions (Na⁺)

Sodium carbonate dissociates in water as follows:
\[ \text{Na₂CO₃} \rightarrow 2 \text{Na}^+ + \text{CO₃}^{2-} \]
From the dissociation equation, we see that for every mole of Na₂CO₃, we get 2 moles of Na⁺. Therefore:
\[ \text{Molarity of Na}^+ = 2 \times \text{Molarity of Na₂CO₃} = 2 \times 0.952 \text{ M} \approx 1.904 \text{ M} \]

### Step 5: Determine the molar concentration of carbonate ions (CO₃²⁻)

From the dissociation equation, we see that for every mole of Na₂CO₃, we get 1 mole of CO₃²⁻. Therefore:
\[ \text{Molarity of CO₃}^{2-} = \text{Molarity of Na₂CO₃} \approx 0.952 \text{ M} \]

### Final Answer

The molar concentrations are:
- Molar concentration of Na⁺: **1.904 M**
- Molar concentration of CO₃²⁻: **0.952 M**