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Heptane and Octane form an ideal solutio...

Heptane and Octane form an ideal solution. At 373 K the vapour pressure of the two liquid components are `105.2" kP"_(a)` and `46.8" kP"_(a)` respectively. What will be the vapour pressure of a mixture of 26.0 g heptane and 35g of octane ?

Text Solution

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Number of moles of octane `n_(0)=("Weight")/("GMwt")=(35)/(114)=0.307`
Number of moles of heptane `n_(s)=("Weight")/("GMwt")=(26)/(100)=0.26`
Mole fraction of octane `X_(0)=(n_(0))/(n_(0)+n_(s))=(0.307)/(0.307+0.26)=0.541`
Mole fraction of heptane `X_(s)=(n_(s))/(n_(0)+n_(s))=(0.26)/(0.307+0.26)=0.459`
Given vapour pressure of heptane `P_(1)=105.2" kP"_(a)`
vapour pressure of Octane `P_(2)=46.8" kP"_(a)`
26 gms of heptane and 35 gms of octane are mixed.
In that mixture
The vapour pressure of heptane `(P_(11))=P_(1)xx X_(s)`
`=105.2xx0.459=48.28" kP"_(a)`
The vapour pressure of Octane `(P_(22))=P_(2)xx X_(0)`
`=46.8xx0.541`
`=25.32" kP"_(a)`
Total pressure of the mixture (P) `=P_(11)+P_(22)`
`=25.32+48.28`
`=73.6" kP"_(a)`
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