Question

a 9.00 l tank at 28.8 °c is filled with 4.37 g of sulfur hexafluoride gas and 2.37 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. round each of your answers to 3 significant digits. sulfur hexafluoride mole fraction: partial pressure: atm chlorine pentafluoride mole fraction: partial pressure: atm total pressure in tank: atm

Explanation:

Step1: Calculate molar masses

Molar mass of SF₆: $32.07 + 6×19.00 = 146.07\ \text{g/mol}$
Molar mass of ClF₅: $35.45 + 5×19.00 = 130.45\ \text{g/mol}$

Step2: Find moles of each gas

Moles of SF₆: $n_{SF₆} = \frac{4.37}{146.07} ≈ 0.0299\ \text{mol}$
Moles of ClF₅: $n_{ClF₅} = \frac{2.37}{130.45} ≈ 0.0182\ \text{mol}$

Step3: Compute total moles

$n_{total} = 0.0299 + 0.0182 = 0.0481\ \text{mol}$

Step4: Calculate mole fractions

$X_{SF₆} = \frac{0.0299}{0.0481} ≈ 0.622$
$X_{ClF₅} = \frac{0.0182}{0.0481} ≈ 0.378$

Step5: Convert temperature to Kelvin

$T = 28.8 + 273.15 = 301.95\ \text{K}$

Step6: Find total pressure (Ideal Gas Law)

$P_{total} = \frac{n_{total}RT}{V} = \frac{0.0481×0.0821×301.95}{9.00} ≈ 0.132\ \text{atm}$

Step7: Calculate partial pressures

$P_{SF₆} = X_{SF₆}×P_{total} ≈ 0.622×0.132 ≈ 0.0821\ \text{atm}$
$P_{ClF₅} = X_{ClF₅}×P_{total} ≈ 0.378×0.132 ≈ 0.0499\ \text{atm}$

Answer:

sulfur hexafluoride mole fraction: 0.622
sulfur hexafluoride partial pressure: 0.0821 atm
chlorine pentafluoride mole fraction: 0.378
chlorine pentafluoride partial pressure: 0.0499 atm
Total pressure in tank: 0.132 atm