Which oen of the following has larger number of molecules ?

Options

A. `11 g CO_(2)`
B. `12 gO_(2)`
C. `2 gH_(2)`
D. `21 gN_(2)`

Answer

To determine which of the given options has the largest number of molecules, we will calculate the number of moles for each substance and then use Avogadro's number to find the number of molecules.

### Step-by-Step Solution:

1. **Identify the Molar Mass of Each Substance:**
- Carbon Dioxide (CO₂): Molar mass = 12 (C) + 2 × 16 (O) = 44 g/mol
- Oxygen (O₂): Molar mass = 2 × 16 (O) = 32 g/mol
- Hydrogen (H₂): Molar mass = 2 × 1 (H) = 2 g/mol
- Nitrogen (N₂): Molar mass = 2 × 14 (N) = 28 g/mol

2. **Calculate the Number of Moles for Each Option:**
- **Option 1: 11 g of CO₂**
\[ \text{Moles of CO₂} = \frac{11 \text{ g}}{44 \text{ g/mol}} = \frac{1}{4} \text{ moles} \]
- **Option 2: 12 g of O₂**
\[ \text{Moles of O₂} = \frac{12 \text{ g}}{32 \text{ g/mol}} = \frac{3}{8} \text{ moles} \]
- **Option 3: 2 g of H₂**
\[ \text{Moles of H₂} = \frac{2 \text{ g}}{2 \text{ g/mol}} = 1 \text{ mole} \]
- **Option 4: 21 g of N₂**
\[ \text{Moles of N₂} = \frac{21 \text{ g}}{28 \text{ g/mol}} = \frac{3}{4} \text{ moles} \]

3. **Calculate the Number of Molecules for Each Option Using Avogadro's Number (Nₐ = \(6.022 \times 10^{23}\) molecules/mol):**
- **Option 1: CO₂**
\[ \text{Number of molecules (N₁)} = \left(\frac{1}{4}\right) \times Nₐ = \frac{Nₐ}{4} \]
- **Option 2: O₂**
\[ \text{Number of molecules (N₂)} = \left(\frac{3}{8}\right) \times Nₐ = \frac{3Nₐ}{8} \]
- **Option 3: H₂**
\[ \text{Number of molecules (N₃)} = 1 \times Nₐ = Nₐ \]
- **Option 4: N₂**
\[ \text{Number of molecules (N₄)} = \left(\frac{3}{4}\right) \times Nₐ = \frac{3Nₐ}{4} \]

4. **Compare the Number of Molecules:**
- N₁ = \( \frac{Nₐ}{4} \)
- N₂ = \( \frac{3Nₐ}{8} \)
- N₃ = \( Nₐ \)
- N₄ = \( \frac{3Nₐ}{4} \)

From the calculations:
- \( N₁ < N₂ < N₄ < N₃ \)

5. **Conclusion:**
The substance with the largest number of molecules is **2 grams of H₂**.

### Final Answer:
**C. 2 grams of H₂ has the largest number of molecules.**