Given the density of air is 1.29 `kg m^(-3)`. Then, the relative density of air is ________.

Answer

To find the relative density of air, we can follow these steps:

### Step 1: Understand the Concept of Relative Density
Relative density (also known as specific gravity) is defined as the ratio of the density of a substance to the density of a reference substance, typically water for liquids and gases.

### Step 2: Identify the Given Values
- Density of air = 1.29 kg/m³
- Density of water = 1000 kg/m³ (or 10³ kg/m³)

### Step 3: Write the Formula for Relative Density
The formula for relative density is:
\[ \text{Relative Density} = \frac{\text{Density of substance}}{\text{Density of reference substance}} \]

### Step 4: Substitute the Values into the Formula
Substituting the values we have:
\[ \text{Relative Density of air} = \frac{\text{Density of air}}{\text{Density of water}} = \frac{1.29 \, \text{kg/m}^3}{1000 \, \text{kg/m}^3} \]

### Step 5: Perform the Calculation
Now, we can perform the calculation:
\[ \text{Relative Density of air} = \frac{1.29}{1000} = 0.00129 \]

### Step 6: Present the Final Answer
The relative density of air is:
\[ \text{Relative Density of air} = 0.00129 \]
Since relative density is a dimensionless quantity, we can express it as:
\[ \text{Relative Density of air} = 1.29 \times 10^{-3} \]

### Final Answer
The relative density of air is **1.29 × 10⁻³** (unitless).

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