`1.6g` `of` `CH_(4)` contains same number of electrons as in :

Options

A. `4.48L` of `CH_(4)at NTP`
B. `6.02xx10^(22) "molecules of" CH_(4)`
C. `1 mol of Na`
D. `2.24 L` of `O_(2) at STP`

Explanation

To solve the problem, we need to determine how many electrons are present in 1.6 grams of methane (CH₄) and then compare that number to the options provided.

### Step-by-Step Solution:

1. **Calculate the Molar Mass of CH₄:**
- The molar mass of carbon (C) = 12 g/mol
- The molar mass of hydrogen (H) = 1 g/mol
- Since CH₄ has one carbon atom and four hydrogen atoms:
\[ \text{Molar mass of CH₄} = 12 + (4 \times 1) = 16 \text{ g/mol} \]

2. **Calculate the Number of Moles of CH₄ in 1.6 g:**
- Use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{1.6 \text{ g}}{16 \text{ g/mol}} = 0.1 \text{ moles} \]

3. **Determine the Total Number of Electrons in CH₄:**
- Each molecule of CH₄ contains:
- 1 carbon atom (6 electrons)
- 4 hydrogen atoms (4 x 1 = 4 electrons)
- Total electrons in one molecule of CH₄:
\[ \text{Total electrons in CH₄} = 6 + 4 = 10 \text{ electrons} \]
- Therefore, the total number of electrons in 0.1 moles of CH₄:
\[ \text{Total electrons} = 10 \text{ electrons/molecule} \times 0.1 \text{ moles} \times 6.02 \times 10^{23} \text{ molecules/mole} = 0.1 \times 6.02 \times 10^{24} = 6.02 \times 10^{23} \text{ electrons} \]

4. **Compare with the Options:**
- We need to check which option has the same number of electrons (6.02 x 10²³).
- **Option 1:** 4.48 liters of CH₄ at NTP
- Moles of CH₄ = \(\frac{4.48 \text{ L}}{22.4 \text{ L/mol}} = 0.2 \text{ moles}\)
- Total electrons = \(10 \times 0.2 \times 6.02 \times 10^{23} = 1.204 \times 10^{24} \text{ electrons}\) (not equal)
- **Option 2:** 6.02 x 10²² molecules
- Moles = \(\frac{6.02 \times 10^{22}}{6.02 \times 10^{23}} = 0.1 \text{ moles}\)
- Total electrons = \(10 \times 0.1 \times 6.02 \times 10^{23} = 6.02 \times 10^{23} \text{ electrons}\) (equal)
- **Option 3:** 1 mole of Na
- Total electrons = \(1 \times 11 \times 6.02 \times 10^{23} = 6.62 \times 10^{24} \text{ electrons}\) (not equal)
- **Option 4:** 2.24 liters of O₂ at NTP
- Moles of O₂ = \(\frac{2.24 \text{ L}}{22.4 \text{ L/mol}} = 0.1 \text{ moles}\)
- Total electrons = \(16 \times 0.1 \times 6.02 \times 10^{23} = 9.632 \times 10^{23} \text{ electrons}\) (not equal)

### Conclusion:
The correct answer is **Option 2: 6.02 x 10²² molecules**.