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At 298K , the vapour pressure of pure li...

At `298K` , the vapour pressure of pure liquid n-butane is 1823 torr and vapour pressure of pure n-pentane is 521 torr and form nearly an ideal solution.
a. Find the total vapour pressure at `298 K` of a liquid solution containing `10%` n-butane and `90%` n-pentane by weight,
b. Find the mole fraction of n-butane in solution exerting a total vapour pressure of 760 torr.
c. What is composition of vapours of two components (mole fraction in vapour state)?

Text Solution

Verified by Experts

The mole fraction in vapour state :
`chi_n^v- "butane" =(0.122xx1823)/(679.84)=0.327`
`chi_n^v-"Pentane" = (0.278xx521)/(679x84)=0.673`
`chi_("vapour")^v` is also called as vapour fraction.
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