The vapour density of a gas is 35.5. The volume occupied by 3.55 g of the gas at S.T.P is?

Options

A. 1.12 litres
B. 11.2 litres
C. 22.4 litres
D. 44.8 litres.

Explanation

To solve the problem of finding the volume occupied by 3.55 g of a gas at STP given its vapor density, we can follow these steps:

### Step-by-step Solution:

1. **Understand the relationship between vapor density and molecular mass:**
- The vapor density (VD) of a gas is defined as half of its molecular mass (M). Therefore, we can express the molecular mass as:
\[ M = 2 \times \text{Vapor Density} \]

2. **Calculate the molecular mass of the gas:**
- Given that the vapor density (VD) is 35.5, we can calculate the molecular mass:
\[ M = 2 \times 35.5 = 71 \text{ g/mol} \]

3. **Use the molar volume at STP:**
- At Standard Temperature and Pressure (STP), one mole of any gas occupies 22.4 liters. Therefore, we can establish the relationship:
\[ 71 \text{ g of gas} \rightarrow 22.4 \text{ L} \]

4. **Determine the volume occupied by 1 gram of the gas:**
- To find the volume occupied by 1 gram of the gas, we divide the molar volume by the molecular mass:
\[ \text{Volume per gram} = \frac{22.4 \text{ L}}{71 \text{ g}} \approx 0.3169 \text{ L/g} \]

5. **Calculate the volume occupied by 3.55 grams of the gas:**
- Now, we can find the volume occupied by 3.55 grams of the gas by multiplying the volume per gram by the mass:
\[ \text{Volume for 3.55 g} = 3.55 \text{ g} \times 0.3169 \text{ L/g} \approx 1.12 \text{ L} \]

### Final Answer:
The volume occupied by 3.55 g of the gas at STP is approximately **1.12 liters**.

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