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A neon-dioxygen mixture contains 70.6 g ...

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 ?

Text Solution

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Number of moles of dioxygen `= ("Mass of" O_(2))/("Molar mass ") = ((70.6 g))/(32.0 gmol^(-1)) = 2.21` mol
Number of moles of neon `= ("Mass of" Ne)/("Molar mass") = ((167.5 g))/((20 gmol^(-1))) = 8.375` mol
Moles of fraction of dioxygen `= ((2.21 mol))/((2.21 + 0.375) mol) = (2.21)/(10.585) = 0.21`
Mole fraction of neon `= 1-0.21 = 0.79`
Partial pressure of dioxygen `(p_(O_(2))) = "Moles fraction" xx "total pressure"`
`= (0.2) xx (25 "bar") = 5.25 "bar"`
Partial pressure of neon `(p_(Ne)) = "Mole fraction" xx "total pressure"`
`(0.79) xx (25 "bar") = 19.75` bar
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