Which of the following contains maximum number of molecules ?

Question

Accepted Answer

To determine which option contains the maximum number of molecules, we need to consider the molecular weight of each substance and use Avogadro's number to calculate the number of molecules in 1 gram of each substance.

1. **Option 1: $1 g H_2$**
   - Molecular weight of $H_2$ = 2 g/mol
   - Number of moles in 1 g of $H_2$ = 1 g / 2 g/mol = 0.5 moles
   - Number of molecules = 0.5 moles × Avogadro's number ($6.022 	imes 10^{23}$ molecules/mol) = $3.011 	imes 10^{23}$ molecules

2. **Option 2: $1 g N_2$**
   - Molecular weight of $N_2$ = 28 g/mol
   - Number of moles in 1 g of $N_2$ = 1 g / 28 g/mol ≈ 0.0357 moles
   - Number of molecules = 0.0357 moles × Avogadro's number = $2.15 	imes 10^{22}$ molecules

3. **Option 3: $1 g CO_2$**
   - Molecular weight of $CO_2$ = 44 g/mol
   - Number of moles in 1 g of $CO_2$ = 1 g / 44 g/mol ≈ 0.0227 moles
   - Number of molecules = 0.0227 moles × Avogadro's number = $1.37 	imes 10^{22}$ molecules

4. **Option 4: $1 g CH_4$**
   - Molecular weight of $CH_4$ = 16 g/mol
   - Number of moles in 1 g of $CH_4$ = 1 g / 16 g/mol = 0.0625 moles
   - Number of molecules = 0.0625 moles × Avogadro's number = $3.76 	imes 10^{22}$ molecules

**Explanation:**

- **Option 1 ($1 g H_2$)** contains the maximum number of molecules because it has the smallest molecular weight (2 g/mol), resulting in the highest number of moles per gram. Consequently, it has the highest number of molecules when calculated using Avogadro's number.

- **Option 2 ($1 g N_2$)**, **Option 3 ($1 g CO_2$)**, and **Option 4 ($1 g CH_4$)** all have higher molecular weights compared to $H_2$, which means fewer moles per gram and thus fewer molecules.

Therefore, based on the calculations, Option 1 is indeed the correct answer as it contains the maximum number of molecules.