How many molecules of water of hydration are present in 630 mg of oxalic acid `(H_(2)C_(2)O_(4).2H_(2)O)` ?

Explanation

To determine how many molecules of water of hydration are present in 630 mg of oxalic acid (H₂C₂O₄·2H₂O), we can follow these steps:

### Step 1: Calculate the molar mass of oxalic acid dihydrate (H₂C₂O₄·2H₂O)

1. **Calculate the molar mass of H₂C₂O₄:**
- Hydrogen (H): 2 × 1.01 g/mol = 2.02 g/mol
- Carbon (C): 2 × 12.01 g/mol = 24.02 g/mol
- Oxygen (O): 4 × 16.00 g/mol = 64.00 g/mol
- Total for H₂C₂O₄ = 2.02 + 24.02 + 64.00 = 90.04 g/mol

2. **Calculate the molar mass of 2H₂O:**
- Water (H₂O): 2 × (2 × 1.01 g/mol + 16.00 g/mol) = 2 × (2.02 + 16.00) = 2 × 18.02 = 36.04 g/mol

3. **Total molar mass of H₂C₂O₄·2H₂O:**
- Total = 90.04 g/mol + 36.04 g/mol = 126.08 g/mol

### Step 2: Convert the mass of oxalic acid to moles

- Given mass = 630 mg = 0.630 g
- Moles of H₂C₂O₄·2H₂O = mass (g) / molar mass (g/mol)
- Moles = 0.630 g / 126.08 g/mol = 0.00499 moles (approximately)

### Step 3: Calculate the number of water molecules

- Each molecule of H₂C₂O₄·2H₂O contains 2 water molecules.
- Therefore, the number of moles of water = 2 × moles of H₂C₂O₄·2H₂O
- Moles of water = 2 × 0.00499 = 0.00998 moles

### Step 4: Convert moles of water to molecules

- Use Avogadro's number (6.022 × 10²³ molecules/mol) to convert moles to molecules.
- Number of water molecules = moles of water × Avogadro's number
- Number of water molecules = 0.00998 moles × 6.022 × 10²³ molecules/mol ≈ 6.01 × 10²¹ molecules

### Final Answer
There are approximately **6.01 × 10²¹ molecules of water of hydration** present in 630 mg of oxalic acid (H₂C₂O₄·2H₂O).

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