Using Graham's Law of Diffusion and the given atomic masses (H=1.0, C=12.0, N=14.0, O=16.0), determine which of the following gases has the lowest rate of diffusion: A. Nitrogen (N2), B. Ammonia (NH3)

• A. Nitrogen ($\text{N}_2$) Molar mass of $\text{N}_2 = 2 \times \text{Atomic mass of N}$ Molar mass of $\text{N}_2 = 2 \times 14.0 = 28.0 \text{ g/mol}$

• B. Ammonia ($\text{NH}_3$) Molar mass of $\text{NH}_3 = \text{Atomic mass of N} + (3 \times \text{Atomic mass of H})$ Molar mass of $\text{NH}_3 = 14.0 + (3 \times 1.0) = 14.0 + 3.0 = 17.0 \text{ g/mol}$

• C. Oxygen ($\text{O}_2$) Molar mass of $\text{O}_2 = 2 \times \text{Atomic mass of O}$ Molar mass of $\text{O}_2 = 2 \times 16.0 = 32.0 \text{ g/mol}$

• D. Methane ($\text{CH}_4$) Molar mass of $\text{CH}_4 = \text{Atomic mass of C} + (4 \times \text{Atomic mass of H})$ Molar mass of $\text{CH}_4 = 12.0 + (4 \times 1.0) = 12.0 + 4.0 = 16.0 \text{ g/mol}$

• Nitrogen ($\text{N}_2$): 28.0 g/mol
• Ammonia ($\text{NH}_3$): 17.0 g/mol
• Oxygen ($\text{O}_2$): 32.0 g/mol
• Methane ($\text{CH}_4$): 16.0 g/mol

Oxygen ($\text{O}_2$) has the highest molar mass (32.0 g/mol) among the given options.

According to Graham's Law, the gas with the highest molar mass will have the lowest rate of diffusion. Therefore, Oxygen ($\text{O}_2$) has the lowest rate of diffusion.

The correct option is C.

The final answer is $\boxed{\text{C}}$