Answer:
- Mole fraction of Chlorine Pentafluoride
= 0.265
- Partial Pressure of Chlorine Pentafluoride
= 16.05 kPa
- Mole fraction of Sulfur Hexafluoride
= 0.735
- Partial Pressure of Sulfur Hexafluoride
= 44.53 kPa
Total Pressure exerted by the gases = 60.58 kPa
Explanation:
First of, we calculate the number of moles of each gas present.
Number of moles = (Mass)/(Molar Mass)
For ClF₅
Mass = 4.28 g
Molar Mass = 130.445 g/mol
number of moles of Chlorine Pentafluoride
= (4.28/130.445) = 0.0328 moles
For SF₆
Mass = 13.3 g
Molar Mass = 146.06 g/mol
number of moles of Sulfur Hexafluoride
= (13.3/146.06) = 0.0911 moles
Total number of moles present = 0.0328 + 0.0911 = 0.1239 moles.
Using the ideal gas equation
PV = nRT
P = total pressure in the tank = ?
V = volume of the tank = 5.00 L = 0.005 m³
R = molar gas constant = 8.314 J/mol.K
T = temperature of the tank = 20.9°C = 294.05 K
n = total number of moles present = 0.1239 moles
P × 0.005 = (0.1239 × 8.314 × 294.05)
P = 60,580.45 Pa = 60.58 kPa.
- Mole fraction of a particular component of interest = (number of moles of the component of interest) ÷ (total number of moles)
- Partial Pressure of a particular component of interest = (mole fraction of that component of interest) × (total pressure)
This is Dalton's law of Partial Pressure.
- Mole fraction of Chlorine Pentafluoride
= (0.0328/0.1239) = 0.265
- Partial Pressure of Chlorine Pentafluoride
= 0.265 × 60.58 = 16.05 kPa
- Mole fraction of Sulfur Hexafluoride
= (0.0911/0.1239) = 0.735
- Partial Pressure of Sulfur Hexafluoride
= 0.735 × 60.58 = 44.53 kPa
Total Pressure exerted by the gases = 16.04 + 44.53 = 60.58 kPa
Hope this Helps!!!
= 0.3571 L or 0.4 L