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Pressure of 0.210 bat at 37^(@)C. The va...

Pressure of `0.210` bat at `37^(@)C`. The vapour pressure of pure liquid `(vapour pressure of water at `37^(@)C`=0.198 bar in bar is

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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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