Answer:
Mass of sample = 8.483 g
Explanation:
Given data;
Volume of mixture = 12.8 L
Pressure = 605.6 mmHg ( 605.6 / 760 = 0.797 atm)
Temperature = 21.6 °C (21.6 + 271.15 = 294.8 K)
Partial pressure of chlorine = 143 mmHg ( 143/760 = 0.19 atm)
Solution:
First of all we will determine the number of moles of 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
partial pressure of chlorine is 0.19 atm so 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
Mass of chlorine = moles × molar mass
Mass of chlorine = 0.1013 mol × 71 g/mol
Mass of chlorine = 7.2 g
Mass of helium = moles × molar mass
Mass of helium = 0.3207 mol × 4 g/mol
Mass of helium = 1.283 g
Mass of sample = mass of chlorine + mass of helium
Mass of sample = 7.2 g + 1.283 g
Mass of sample = 8.483 g
