40 g of caustic soda contain :

Options

A. `6.02 xx 10^(23)` atoms of H
B. 22.4 litres of `N_(2)`
C. `6.02 xx 10^(24)` molecules of `O_(2)`
D. 4 g of Na

Explanation

To determine the composition of 40 g of caustic soda (sodium hydroxide, NaOH), we will follow these steps:

### Step 1: Calculate the Molar Mass of NaOH
The molar mass of NaOH is calculated by adding the atomic weights of its constituent elements:
- Sodium (Na): 23 g/mol
- Oxygen (O): 16 g/mol
- Hydrogen (H): 1 g/mol

**Molar Mass of NaOH = 23 + 16 + 1 = 40 g/mol**

### Step 2: Calculate the Number of Moles in 40 g of NaOH
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{40 \text{ g}}{40 \text{ g/mol}} = 1 \text{ mole} \]

### Step 3: Determine the Number of Atoms in 1 Mole of NaOH
1 mole of NaOH contains:
- 1 mole of Sodium (Na)
- 1 mole of Oxygen (O)
- 1 mole of Hydrogen (H)

### Step 4: Calculate the Number of Atoms of Each Element
Using Avogadro's number (approximately \(6.02 \times 10^{23}\) atoms/mole):
- Number of Sodium atoms = \(1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mole} = 6.02 \times 10^{23} \text{ atoms}\)
- Number of Oxygen atoms = \(1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mole} = 6.02 \times 10^{23} \text{ atoms}\)
- Number of Hydrogen atoms = \(1 \text{ mole} \times 6.02 \times 10^{23} \text{ atoms/mole} = 6.02 \times 10^{23} \text{ atoms}\)

### Step 5: Analyze the Given Options
1. **6.02 x 10^23 atoms of Hydrogen**: This is correct as we calculated 1 mole of Hydrogen, which equals \(6.02 \times 10^{23}\) atoms.
2. **22.4 liters of N2**: Incorrect, as NaOH does not contain nitrogen.
3. **6.02 x 10^24 molecules of O2**: Incorrect, as we only have 1 mole of O in NaOH, which does not relate to O2.
4. **Moles of Na**: Incorrect, as we already calculated 1 mole of Na.

### Conclusion
The correct answer is that 40 g of caustic soda contains **6.02 x 10^23 atoms of Hydrogen**.

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