If 11.0g of a gas occupies 5.6 dm$^3$ at s.t.p., calculate its vapour density (1 mole of a gas occupies 22.4 dm$^3$).

1. Analyze the Question

We are given the mass of a gas (11.0g) and the volume it occupies at standard temperature and pressure (s.t.p.) (5.6 dm³). We are also given the molar volume of a gas (22.4 dm³). We need to find the vapour density of the gas.

2. Explain the Concept

• Molar Mass ($M$): This is the mass of one mole of a substance. Since 1 mole of any gas occupies 22.4 dm³ at s.t.p., we can find the molar mass using the ratio: $$\text{Molar Mass} = \frac{\text{Mass of gas} \times 22.4 \text{ dm}^3}{\text{Volume of gas}}$$
• Vapour Density (V.D.): This is defined as the ratio of the mass of a certain volume of a gas to the mass of an equal volume of hydrogen gas under the same conditions. The relationship between Molar Mass and Vapour Density is: $$\text{Molar Mass} = 2 \times \text{Vapour Density}$$ Therefore, Vapour Density = $\frac{\text{Molar Mass}}{2}$

3. Show the Solution

Step 1: Calculate the Molar Mass of the gas. Using the given values:

• Mass = 11.0g
• Volume = 5.6 dm³
• Molar Volume = 22.4 dm³

Step 2: Calculate the Vapour Density.

4. Identify the Answer

Comparing our result to the options provided, the correct answer is A.

5. Tip

Always remember the relationship Molar Mass = 2 × Vapour Density. A common mistake students make is forgetting to divide the molar mass by 2 at the end, or confusing molar volume (22.4) with the constant used for vapour density (2). Keep these constants clear in your mind!