Questions: A 8.00 L tank at -9.5°C is filled with 15.0 g of chlorine pentafluoride gas and 8.06 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction of each gas. Be sure each of your answer entries has the correct number of significant digits.

Solution Steps

First, we need to convert the mass of each gas to moles using their molar masses.

• Molar mass of chlorine pentafluoride (ClF\(_5\)): \[ M_{\text{ClF}_5} = 35.453 + 5 \times 18.998 = 125.443 \, \text{g/mol} \]
• Molar mass of sulfur hexafluoride (SF\(_6\)): \[ M_{\text{SF}_6} = 32.065 + 6 \times 18.998 = 146.065 \, \text{g/mol} \]

Now, calculate the moles of each gas:

• Moles of ClF\(_5\): \[ n_{\text{ClF}_5} = \frac{15.0 \, \text{g}}{125.443 \, \text{g/mol}} = 0.1196 \, \text{mol} \]
• Moles of SF\(_6\): \[ n_{\text{SF}_6} = \frac{8.06 \, \text{g}}{146.065 \, \text{g/mol}} = 0.0552 \, \text{mol} \]

Next, we calculate the total number of moles of gas in the tank: \[ n_{\text{total}} = n_{\text{ClF}_5} + n_{\text{SF}_6} = 0.1196 + 0.0552 = 0.1748 \, \text{mol} \]

Finally, we calculate the mole fraction of each gas:

• Mole fraction of ClF\(_5\): \[ x_{\text{ClF}_5} = \frac{n_{\text{ClF}_5}}{n_{\text{total}}} = \frac{0.1196}{0.1748} = 0.6843 \]
• Mole fraction of SF\(_6\): \[ x_{\text{SF}_6} = \frac{n_{\text{SF}_6}}{n_{\text{total}}} = \frac{0.0552}{0.1748} = 0.3157 \]

Final Answer