A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of the gases in the cylinder is 25 bar.What are the partial pressures of dioxygen and neon in the mixture ?
(3 marks)
Answers
Explanation:
Given: mass of neon = 167.5 g, mass of O₂ = 70.6 g , Ptotal = 25 bar
moles of Ne = 165.5/32 = 8.38 moles
moles of O₂ = 70.6/32 = 2.21 moles
Total moles = 8.38 + 2.21 = 10.59 moles
Mole fraction of Ne = moles of Ne/total moels
= 8.38/10.59
= 0.79
mole fraction of O₂ = 2.21/10.59
= 0.21
Partial pressure of O₂ = mole fraction × Ptotal
= 0.21 × 25
= 5.25 bar
Partial pressure of Ne = 0.79 × 25
= 19.75 bar
Answer:
The mass of dioxygen is 70.6 g.
The molar mass of dioxygen is 32 g/mol.
The number of moles of dioxygen
=
32 g/mol
70.6 g
= 2.21 mol
The mass of neon is 167.5 g.
The molar mass of neon is 20.2 g/mol.
The number of moles of neon
=
20.2 g/mol
167.5 g
= 8.29 mol
The total number of moles of neon and dioxygen = 2.21 + 8.29= 10.49
The mole fraction of neon =
10.49
8.29
= 0.789
Total pressure in the cylinder is 25 bar.
The partial pressure of neon = 0.789 × 25 bar= 19.7 bar
The partial pressure of dioxygen = 25 bar − 19.7 bar = 5.3 bar
Hope its correct.
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