A gas has a vapour density 1.2. The volume occupied by gram of the gas at STP will be:

Options

A. 1L
B. 9L
C. 10L
D. 11.2L

Answer

To solve the problem, we need to find the volume occupied by 1 gram of a gas with a vapor density of 1.2 at Standard Temperature and Pressure (STP). Here’s a step-by-step solution:

### Step 1: Understand the Concept of Vapor Density
Vapor density 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 of a gas can be calculated using the formula:

\[ \text{Molar Mass} = 2 \times \text{Vapor Density} \]

### Step 2: Calculate the Molar Mass
Given the vapor density of the gas is 1.2, we can calculate the molar mass:

\[ \text{Molar Mass} = 2 \times 1.2 = 2.4 \, \text{g/mol} \]

### Step 3: Calculate the 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:

\[ \text{Number of Moles} = \frac{1 \, \text{g}}{2.4 \, \text{g/mol}} = \frac{1}{2.4} \, \text{mol} \]

### Step 4: Calculate the Volume at STP
At STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, the volume occupied by \(\frac{1}{2.4}\) moles of gas can be calculated as follows:

\[ \text{Volume} = \text{Number of Moles} \times 22.4 \, \text{L/mol} \]

Substituting the number of moles:

\[ \text{Volume} = \left(\frac{1}{2.4}\right) \times 22.4 \, \text{L} = \frac{22.4}{2.4} \, \text{L} \]

Calculating this gives:

\[ \text{Volume} = 9.33 \, \text{L} \]

### Final Answer
The volume occupied by 1 gram of the gas at STP is approximately **9.33 liters**.

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