Answer:
The sample's mass totals 8.483 g
Clarification:
Provided data includes;
Mixture volume = 12.8 L
Pressure = 605.6 mmHg ( 605.6 / 760 = 0.797 atm)
Temperature = 21.6 °C (21.6 + 271.15 = 294.8 K)
Chlorine's partial pressure = 143 mmHg ( 143/760 = 0.19 atm)
Procedure:
Initially, we will figure out the number of moles in the mixture.
PV = nRT
n = PV/RT
n = 0.797atm × 12.8L / 0.0821 atm. dm³ mol⁻¹ K⁻¹ ×294.8 K
n = 10.202 / 24.2031
n = 0.422 mol
Given the chlorine's partial pressure of 0.19 atm, the mole fraction is
mole fraction = 0.19/0.797
mole fraction = 0.24
Moles of chlorine = 0.24 × 0.422 = 0.1013 mol
Moles of helium = moles of mixture - moles of chlorine
Moles of helium = 0.422 - 0.1013
Moles of helium = 0.3207 mol
Chlorine mass = moles × molar mass
Chlorine mass = 0.1013 mol × 71 g/mol
Chlorine mass = 7.2 g
Helium mass = moles × molar mass
Helium mass = 0.3207 mol × 4 g/mol
Helium mass = 1.283 g
Total sample mass = mass of chlorine + mass of helium
Total sample mass = 7.2 g + 1.283 g
Total sample mass = 8.483 g
