The vapour density of a mixture containing equal number of moles of methane and ethane at `STP` is

Options

A. `11.5`
B. `11.0`
C. 23
D. `12.0`

Explanation

To find the vapor density of a mixture containing equal numbers of moles of methane and ethane at STP, we can follow these steps:

### Step 1: Determine the Molecular Weights
- The molecular weight of methane (CH₄) is 16 g/mol.
- The molecular weight of ethane (C₂H₆) is 30 g/mol.

### Step 2: Assume Equal Moles
- Let’s assume we have \( x \) moles of methane and \( x \) moles of ethane. Since the question states that there are equal numbers of moles of both gases, we can proceed with this assumption.

### Step 3: Calculate Total Mass of the Mixture
- The total mass of the mixture can be calculated as follows:
\[ \text{Total mass} = (\text{moles of methane} \times \text{molecular weight of methane}) + (\text{moles of ethane} \times \text{molecular weight of ethane}) \]
\[ \text{Total mass} = (x \times 16) + (x \times 30) = 16x + 30x = 46x \]

### Step 4: Calculate the Average Molecular Weight of the Mixture
- The average molecular weight (M) of the mixture can be calculated using the formula:
\[ M = \frac{(\text{moles of methane} \times \text{molecular weight of methane}) + (\text{moles of ethane} \times \text{molecular weight of ethane})}{\text{total moles}} \]
\[ M = \frac{(x \times 16) + (x \times 30)}{x + x} = \frac{46x}{2x} = 23 \text{ g/mol} \]

### Step 5: Calculate the Vapor Density
- The vapor density (VD) is defined as half of the molecular weight:
\[ \text{Vapor Density} = \frac{M}{2} = \frac{23}{2} = 11.5 \]

### Final Answer
The vapor density of the mixture of methane and ethane at STP is **11.5**.

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