Question

Which of the following gas is lighter than air?

• A. Oxygen
• B. Ammonia
• C. Chlorine
• D. Carbon Dioxide

Answer 1

The correct option is B

Ammonia

The explanation for the correct option:

Option (B) Ammonia

• To check for the gas lighter than air, calculate the vapor density of all these gases and compare them with the vapor density of air.

Vapor density

The weight of a vapor or gas compared to an equal volume of air is known as vapor density.

Step 1: Given data

The following gases are given: Oxygen, Ammonia, Chlorine, and Carbon dioxide.

The vapor density of air is 14.4.

Step 2: Formula used

The vapor density can be calculated using the formula: $\mathrm{vapour}\mathrm{density}=\frac{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{gas}}{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}}$

Step 3: Calculation of vapor density of ammonia

The vapor density of ammonia is calculated using the formula: $\mathrm{vapour}\mathrm{density}=\frac{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{gas}}{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}}$

Mass of 1 mole of ammonia = 17g

$\mathrm{vapour}\mathrm{density}\mathrm{of}\mathrm{ammonia}=\frac{17}{2}=8.5$

Hence, ammonia's vapor density is less than air's vapor density. Therefore ammonia is lighter than air.

The explanation for incorrect options:

Option A: Oxygen

Step 1: Given data

The following gases are given: Oxygen, Ammonia, Chlorine, and Carbon dioxide.

The vapor density of air is 14.4.

Step 2: Formula used

The vapor density can be calculated using the formula: $\mathrm{vapour}\mathrm{density}=\frac{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{gas}}{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}}$

Step 3: Calculation of vapor density of oxygen

The vapor density of oxygen is calculated using the formula: $\mathrm{vapour}\mathrm{density}=\frac{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{gas}}{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}}$

Mass of 1 mole of oxygen= 32g

$\mathrm{vapour}\mathrm{density}\mathrm{of}\mathrm{oxygen}=\frac{32}{2}=16$

As the vapor density of oxygen is more than that of air, it is heavier than air.

Hence, this option is incorrect.

Option C: Chlorine

Step 1: Given data

The following gases are given: Oxygen, Ammonia, Chlorine, and Carbon dioxide.

The vapor density of air is 14.4.

Step 2: Formula used

The vapor density can be calculated using the formula: $\mathrm{vapour}\mathrm{density}=\frac{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{gas}}{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}}$

Step 3: Calculation of vapor density of chlorine

The vapor density of chlorine is calculated using the formula: $\mathrm{vapour}\mathrm{density}=\frac{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{gas}}{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}}$

Mass of 1 mole of chlorine = 70g

$\mathrm{vapour}\mathrm{density}\mathrm{of}\mathrm{chlorine}=\frac{70}{2}=35.5$

Hence, this option is incorrect.

Option D: Carbon dioxide

Step 1: Given data

The following gases are given: Oxygen, Ammonia, Chlorine, and Carbon dioxide.

The vapor density of air is 14.4.

Step 2: Formula used

The vapor density can be calculated using the formula: $\mathrm{vapour}\mathrm{density}=\frac{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{gas}}{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}}$

Step 3: Calculation of vapor density of carbon dioxide

The vapor density of dioxide is calculated using the formula: $\mathrm{vapour}\mathrm{density}=\frac{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{gas}}{\mathrm{mass}\mathrm{of}1\mathrm{mole}\mathrm{of}\mathrm{hydrogen}}$

Mass of 1 mole of carbon dioxide = 44g

$\mathrm{vapour}\mathrm{density}\mathrm{of}\mathrm{carbon}\mathrm{dioxide}=\frac{44}{2}=22$

Hence, this option is incorrect.

Hence, option B, ammonia is the correct option.