https://file.aiquickdraw.com/mathbot/file/1727546534412e5ncj0c4.jpg

To solve this problem, we will use the **Ideal Gas Law**:

\[
PV = nRT
\]

Where:
- \( P \) is the pressure (in atm),
- \( V \) is the volume (in L),
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant \((0.0821 \, \text{L·atm/mol·K})\),
- \( T \) is the temperature in Kelvin.

Let's break this into steps:

### Step 1: Convert the temperature to Kelvin
Given temperature in Celsius is \(14.9 \, ^\circ \text{C}\):
\[
T = 14.9 + 273.15 = 288.05 \, \text{K}
\]

### Step 2: Calculate the moles of each gas
1. **Sulfur hexafluoride (SF₆)**:
   - Molar mass of SF₆ = 32.06 (S) + 6 × 18.998 (F) = **146.06 g/mol**.
   - Given mass = 18.9 g.
   - Moles of SF₆:
   \[
   n_{\text{SF}_6} = \frac{18.9}{146.06} \approx 0.129 \, \text{mol}
   \]

2. **Chlorine pentafluoride (ClF₅)**:
   - Molar mass of ClF₅ = 35.453 (Cl) + 5 × 18.998 (F) = **130.443 g/mol**.
   - Given mass = 6.41 g.
   - Moles of ClF₅:
   \[
   n_{\text{ClF}_5} = \frac{6.41}{130.443} \approx 0.0491 \, \text{mol}
   \]

### Step 3: Calculate the total number of moles
\[
n_{\text{total}} = n_{\text{SF}_6} + n_{\text{ClF}_5} = 0.129 + 0.0491 = 0.1781 \, \text{mol}
\]

### Step 4: Calculate the mole fractions
1. Mole fraction of SF₆:
   \[
   \chi_{\text{SF}_6} = \frac{n_{\text{SF}_6}}{n_{\text{total}}} = \frac{0.129}{0.1781} \approx 0.724
   \]
2. Mole fraction of ClF₅:
   \[
   \chi_{\text{ClF}_5} = \frac{n_{\text{ClF}_5}}{n_{\text{total}}} = \frac{0.0491}{0.1781} \approx 0.276
   \]

### Step 5: Calculate the total pressure in the tank
Using the Ideal Gas Law to find the total pressure:

\[
P_{\text{total}} = \frac{n_{\text{total}}RT}{V}
\]
- \( V = 5.00 \, \text{L} \)
- \( R = 0.0821 \, \text{L·atm/mol·K} \)
- \( T = 288.05 \, \text{K} \)

\[
P_{\text{total}} = \frac{0.1781 \times 0.0821 \times 288.05}{5.00} \approx 0.840 \, \text{atm}
\]

### Step 6: Calculate the partial pressures
1. Partial pressure of SF₆:
   \[
   P_{\text{SF}_6} = \chi_{\text{SF}_6} \times P_{\text{total}} = 0.724 \times 0.840 \approx 0.609 \, \text{atm}
   \]

2. Partial pressure of ClF₅:
   \[
   P_{\text{ClF}_5} = \chi_{\text{ClF}_5} \times P_{\text{total}} = 0.276 \times 0.840 \approx 0.232 \, \text{atm}
   \]

### Final Answers:
- Mole fraction of SF₆: **0.724**
- Partial pressure of SF₆: **0.609 atm**
- Mole fraction of ClF₅: **0.276**
- Partial pressure of ClF₅: **0.232 atm**
- Total pressure: **0.840 atm**

---

Let me know if you want more details or have any questions! Here are 5 related questions for practice:

1. How would the partial pressures change if the temperature increased?
2. What would happen to the mole fraction of SF₆ if the mass of ClF₅ doubled?
3. Can the Ideal Gas Law be used for real gases at high pressures? Why or why not?
4. How do intermolecular forces affect gas behavior in non-ideal conditions?
5. What happens to the total pressure if the volume of the container is halved?

**Tip:** Always convert temperatures to Kelvin when using the Ideal Gas Law!

A 5.00 L tank at 14.9°C is filled with 18.9 g of sulfur hexafluoride gas and 6.41 g of chlorine pentafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank.

Learn how to calculate the mole fractions and partial pressures of sulfur hexafluoride (SF₆) and chlorine pentafluoride (ClF₅) in a 5.00 L tank at 14.9°C. This problem uses the Ideal Gas Law and involves converting masses to moles, finding mole fractions, and applying the formula for partial pressures.

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