1.

A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture ?

Answer» Number of moles of dioxygen
`=(70.6g)/(32g mol^(-1))`
=2.21 mol
Number of moles of neon
`=(167.5g)/(20g mol^(-1))`
=8.375 mol
Mole fraction of dioxygen
`=(2.21)/(2.21+8.3745)`
`=(2.21)/(10.585)`
=0.21
Mole fraction of neon `=(8.375)/(2.21+8.375)`
`" " =0.79`
Alternatively,
mole fraction of neon =1-0.21=0.79
Partial pressure mole fraction `xx` of a gas total pressure
`rArr"Partial pressure" =0.21xx(25"bar")`
of oxygen =5.25 bar
Partial pressure =0.79 `xx` (25 bar)
of neon =19.75 bar


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