To solve this using Graham's Law of Effusion, we use the following relationship:

$\frac{\text{Rate of effusion of gas A}}{\text{Rate of effusion of gas B}} = \sqrt{\frac{\text{Molar mass of gas B}}{\text{Molar mass of gas A}}}$

Where:

• Gas A is the unknown gas.
• Gas B is oxygen ($O_2$).
• The molar mass of $O_2$ is $32.00 \, \text{g/mol}$.
• The rates of effusion are given: $\text{Rate of effusion of gas A} = 13.45 \, \text{mL/min}, \quad \text{Rate of effusion of gas B} = 8.96 \, \text{mL/min}.$

Step 1: Rearrange Graham's Law to solve for the molar mass of the unknown gas:

$\frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} = \sqrt{\frac{\text{Molar mass of } O_2}{\text{Molar mass of unknown gas}}}$ Square both sides: $\left(\frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}}\right)^2 = \frac{\text{Molar mass of } O_2}{\text{Molar mass of unknown gas}}$ Rearranging to isolate the molar mass of the unknown gas: $\text{Molar mass of unknown gas} = \frac{\text{Molar mass of } O_2}{\left(\frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}}\right)^2}$

Step 2: Calculate the ratio of the rates:

$\frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} = \frac{13.45}{8.96} \approx 1.502$

Step 3: Square the ratio:

$\left(\frac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}}\right)^2 = 1.502^2 \approx 2.256$

Step 4: Solve for the molar mass of the unknown gas:

$\text{Molar mass of unknown gas} = \frac{32.00}{2.256} \approx 14.18 \, \text{g/mol}$

Step 5: Identify the unknown gas:

A gas with a molar mass of approximately $14.18 \, \text{g/mol}$ is most likely nitrogen ($N_2$), which has a molar mass of $14.01 \, \text{g/mol}$.

Final Answer:

The unknown gas is nitrogen ($N_2$).

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Would you like me to explain any specific step in more detail? 😊

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5 Related Questions:

1. What assumptions are made in Graham's law about the gases involved?
2. How would the answer change if the conditions were not identical for both gases?
3. Can Graham's law be applied to gas mixtures? Why or why not?
4. How does the molar mass of a gas affect its rate of diffusion or effusion?
5. What is the kinetic theory explanation for Graham's law?

Tip:

Always ensure that the temperatures and pressures of the gases are identical before applying Graham's law, as deviations can introduce errors in your calculation.