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Question

A mixture containing 1.6 g of O2, 1.4 g of N2 and 0.4 g of He occupies a volume of 10 L at 27C. Calculate the partial pressure of each component of the mixture.


A

PN2=0.500 atm
PO2=0.25 atm
PHe=0.25 atm

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B

PN2=0.246 atm
PO2=0.246 atm
PHe=0.492 atm

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C

PN2=0.123 atm
PO2=0.123 atm
PHe=0.246 atm

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D

PN2=1 atm
PO2=1.123 atm
PHe=2.246 atm

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Solution

The correct option is C

PN2=0.123 atm
PO2=0.123 atm
PHe=0.246 atm


PHe = 0.246 atm, PO2= PN2 = 0.123 atm

Given that V=10 L , T = 27C=300 K
The number of moles are :-

nHe=0.44=0.1
nN2 =1.428= 0.05
nO2 = 1.632 = 0.05

therefore , the total number of moles = 0.1+0.05+0.05 = 0.2

Using the ideal gas equation PV = nRT P×10=0.2×0.082×300P=0.492 atm

As partial pressure = total pressure × mole fraction , we have

PHe = 0.492×0.10.2 = 0.246 atm

PN2= 0.492×0.050.2=0.123 atm

PO2= 0.492×0.050.2 = 0.123 atm


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