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At 80^@C, the vapour pressure of pure li...

At `80^@C`, the vapour pressure of pure liquid `A` is `520 mm` Hg and that of pure liquid `B` is `1000 mm Hg`. If a mixture of solution `A` and `B` boils at `80@C` and `1 atm` pressure, the amount of `A` in the mixture is `(1 atm =760 mm Hg)`
a. `50 mol %` , b.`52 mol %` ,c.`34 mol%` ,d.`48 mol %`

A

0.448

B

44.8

C

0.224

D

2.24

Text Solution

Verified by Experts

The correct Answer is:
A
Mole fraction of liquid `A`, `chi_(l)=0.2/(0.2 + 4.0)=0.048`
Mole fraction of water, `chi_(w)=1.0-0.048=0.952`
`0.210=P_(t)^(@)chi_(l) + P_(w)^(@)chi_(w)`
`0.210=P_(l)^(@) xx 0.048 + 0.198 xx 0.952`
`P_(l)^(@)=(0.210-0.198 xx 0.952)/0.048 =0.448 "bar"`
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