`4*0` g of caustic soda (Mol. Mass 40) contains same number of sodium ions as are present in

Options

A. `10*6` g of `Na_(2)CO_(3)` (Mol. mass 106)
B. `58*5` g of `NaCl` (Formula mass `58*5`)
C. 100 mL of `0*5MNa_(2) SO_(4)` (Formula mass = 142)
D. 1 g-equivalent of `NaNO_(3)` (Mol. mass 85)

Explanation

To solve the problem, we need to determine how many sodium ions are present in 4 grams of caustic soda (NaOH) and then compare that to the sodium ions present in the given options.

### Step-by-Step Solution:

1. **Determine the number of moles of NaOH:**
- The formula to calculate the number of moles is:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
- Here, the given mass of NaOH is 4 g and the molar mass is 40 g/mol.
- Therefore:
\[ \text{Number of moles of NaOH} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ moles} \]

2. **Calculate the number of sodium ions:**
- Each molecule of NaOH produces one sodium ion (Na⁺).
- Thus, the number of sodium ions can be calculated using Avogadro's number (approximately \(6.022 \times 10^{23}\) ions/mol):
\[ \text{Number of sodium ions} = \text{Number of moles of NaOH} \times \text{Avogadro's number} \]
- Therefore:
\[ \text{Number of sodium ions} = 0.1 \text{ moles} \times 6.022 \times 10^{23} \text{ ions/mol} = 6.022 \times 10^{22} \text{ sodium ions} \]

3. **Compare with the options:**
- We need to check which of the given options has the same number of sodium ions (6.022 × 10²²).
- Let's analyze each option:

**Option A: 10.6 g of Na₂CO₃**
- Molar mass of Na₂CO₃ = 106 g/mol.
- Moles of Na₂CO₃:
\[ \text{Moles} = \frac{10.6 \text{ g}}{106 \text{ g/mol}} = 0.1 \text{ moles} \]
- Sodium ions from Na₂CO₃ (2 Na⁺ per formula):
\[ \text{Sodium ions} = 0.1 \text{ moles} \times 2 = 0.2 \text{ moles} = 0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23} \text{ sodium ions} \]
- This is incorrect.

**Option B: 58.5 g of NaCl**
- Molar mass of NaCl = 58.5 g/mol.
- Moles of NaCl:
\[ \text{Moles} = \frac{58.5 \text{ g}}{58.5 \text{ g/mol}} = 1 \text{ mole} \]
- Sodium ions from NaCl (1 Na⁺ per formula):
\[ \text{Sodium ions} = 1 \text{ mole} = 6.022 \times 10^{23} \text{ sodium ions} \]
- This is incorrect.

**Option C: 0.5 M Na₂SO₄ in 100 mL**
- Molarity = 0.5 M, Volume = 100 mL = 0.1 L.
- Moles of Na₂SO₄:
\[ \text{Moles} = 0.5 \text{ moles/L} \times 0.1 \text{ L} = 0.05 \text{ moles} \]
- Sodium ions from Na₂SO₄ (2 Na⁺ per formula):
\[ \text{Sodium ions} = 0.05 \text{ moles} \times 2 = 0.1 \text{ moles} = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22} \text{ sodium ions} \]
- This is correct.

**Option D: 1 gram equivalent of Na₃PO₄**
- Equivalent weight of Na₃PO₄ = 1/3 of molar mass (3 Na⁺).
- This option will also yield sodium ions that do not match.

4. **Conclusion:**
- The correct answer is Option C, which contains the same number of sodium ions as 4 g of caustic soda.