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To understand the density of different oxygen allotropes, knowing their molar masses is essential. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

For dioxygen \(( \text{O}_2 \)), each oxygen atom has an atomic mass of about 16 g/mol. Since there are two oxygen atoms in dioxygen, the molar mass of \( \text{O}_2 \) is:
\[ M_{\text{O}_2} = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol} \]
For ozone \( \text{O}_3 \), with three oxygen atoms, the molar mass is:
\[ M_{\text{O}_3} = 3 \times 16 \text{ g/mol} = 48 \text{ g/mol} \]
These molar masses help us move to the next steps in calculating densities and understanding their relative differences.

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. The equation is:

\[ PV = nRT \]
Here:

• P is the pressure of the gas (in atm).
• V is the volume of the gas (in L).
• n is the number of moles of the gas.
• R is the ideal gas constant (0.0821 L atm/mol K).
• T is the temperature (in K).

Given that we know the pressure (1 atm) and temperature (273.15 K), we can rearrange the Ideal Gas Law equation to solve for volume (V):
\[ V = \frac{nRT}{P} \]
For simplicity, let’s assume we are calculating for 1 mole of each gas. This approach will give us the volume occupied by 1 mole of dioxygen and 1 mole of ozone under standard conditions. Knowing the volume is crucial to calculate the density.

Density is defined as the mass per unit volume of a substance. For gases, it's typically expressed in grams per liter (g/L). Using the mass (m) and volume (V) from previous steps, density (d) can be calculated as:

\[ d = \frac{m}{V} \]
For dioxygen \(\text{O}_2\), with a molar mass of 32 g and assuming the volume for 1 mole from the Ideal Gas Law:
\[ d_{\text{O}_2} = \frac{32 \text{g}}{22.4 \text{L}} = 1.429 \text{ g/L} \]
Similarly, for ozone \(\text{O}_3\), with a molar mass of 48 g and the same volume:

\[ d_{\text{O}_3} = \frac{48 \text{g}}{22.4 \text{L}} = 2.143 \text{ g/L} \]
Finally, we can find the ratio of densities to understand how much heavier ozone is compared to dioxygen:
\[ \frac{d_{\text{O}_3}}{d_{\text{O}_2}} = \frac{2.143 \text{ g/L}}{1.429 \text{ g/L}} \approx 1.5 \]
This ratio indicates that ozone is about 1.5 times denser than dioxygen, showing that it is heavier and could have different physical and chemical behaviors.