11.2 L of `CH_(4)` and 22.4 L of `C_(2)H_(6)` at STP are mixed. Then choose correct statement/ statements

Options

A. Vapour density of the mixture is 12.67
B. Average molecular wt. will be less than 16
C. Average molecular wt. will be greater than 16 and less than 30
D. Average molecular wt. will be greater than 30

Answer

To solve the problem of mixing 11.2 L of \( CH_4 \) and 22.4 L of \( C_2H_6 \) at STP, we will follow these steps:

### Step 1: Calculate the number of moles of \( CH_4 \)
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.

Given:
- Volume of \( CH_4 = 11.2 \, L \)

Using the formula:
\[ \text{Number of moles} = \frac{\text{Volume}}{22.4 \, L} \]
\[ \text{Number of moles of } CH_4 = \frac{11.2 \, L}{22.4 \, L} = 0.5 \, moles \]

### Step 2: Calculate the mass of \( CH_4 \)
The molecular weight of \( CH_4 \) (methane) is 16 g/mol.

Using the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molecular weight} \]
\[ \text{Mass of } CH_4 = 0.5 \, moles \times 16 \, g/mol = 8 \, g \]

### Step 3: Calculate the number of moles of \( C_2H_6 \)
Given:
- Volume of \( C_2H_6 = 22.4 \, L \)

Using the same formula:
\[ \text{Number of moles of } C_2H_6 = \frac{22.4 \, L}{22.4 \, L} = 1 \, mole \]

### Step 4: Calculate the mass of \( C_2H_6 \)
The molecular weight of \( C_2H_6 \) (ethane) is 30 g/mol.

Using the formula:
\[ \text{Mass of } C_2H_6 = 1 \, mole \times 30 \, g/mol = 30 \, g \]

### Step 5: Calculate the total mass and total moles
Total mass:
\[ \text{Total mass} = \text{Mass of } CH_4 + \text{Mass of } C_2H_6 = 8 \, g + 30 \, g = 38 \, g \]

Total moles:
\[ \text{Total moles} = \text{Moles of } CH_4 + \text{Moles of } C_2H_6 = 0.5 + 1 = 1.5 \, moles \]

### Step 6: Calculate the average molecular weight
Using the formula:
\[ M_{\text{average}} = \frac{\text{Total mass}}{\text{Total moles}} = \frac{38 \, g}{1.5 \, moles} = 25.33 \, g/mol \]

### Step 7: Calculate the vapor density
Vapor density is calculated using the formula:
\[ \text{Vapor density} = \frac{M_{\text{average}}}{2} = \frac{25.33 \, g/mol}{2} = 12.67 \]

### Conclusion
Now we can analyze the statements provided:
1. The vapor density of the mixture is 12.67 (Correct).
2. The average molecular weight will be less than 16 (Incorrect, it is 25.33).
3. The average molecular weight will be greater than 16 and less than 30 (Correct, it is 25.33).
4. The average molecular weight will be greater than 30 (Incorrect).

Thus, the correct statements are:
- The vapor density of the mixture is 12.67.
- The average molecular weight will be greater than 16 and less than 30.