A gas has a vapour density 11.2. The volume occupied by gram of the gas at STP will be:

Options

A. 11.2L
B. 22.4L
C. 1L
D. 10L

Explanation

To solve the problem, we need to find the volume occupied by 1 gram of a gas with a vapor density of 11.2 at Standard Temperature and Pressure (STP). Here’s a step-by-step solution:

### Step 1: Understand Vapor Density
Vapor density (VD) is defined as the mass of a certain volume of gas compared to the mass of an equal volume of hydrogen at the same temperature and pressure. The molar mass (M) of a gas can be calculated using the formula:
\[ \text{Molar Mass} = 2 \times \text{Vapor Density} \]

### Step 2: Calculate Molar Mass
Given the vapor density (VD) is 11.2, we can calculate the molar mass (M):
\[ M = 2 \times 11.2 = 22.4 \, \text{g/mol} \]

### Step 3: Calculate Number of Moles
Next, we calculate the number of moles of the gas using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Here, the mass of the gas is 1 gram:
\[ \text{Number of moles} = \frac{1 \, \text{g}}{22.4 \, \text{g/mol}} = \frac{1}{22.4} \, \text{mol} \]

### Step 4: Calculate Volume at STP
At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters. Therefore, the volume occupied by \(\frac{1}{22.4}\) moles of the gas can be calculated as follows:
\[ \text{Volume} = \text{Number of moles} \times 22.4 \, \text{L/mol} \]
\[ \text{Volume} = \left(\frac{1}{22.4}\right) \times 22.4 \, \text{L} = 1 \, \text{L} \]

### Final Answer
The volume occupied by 1 gram of the gas at STP is **1 liter**.

---