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Question

At 80C, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole per cent which at 80C is in equilibrium with the vapour containing 30 mole per cent of benzene.

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Solution

PoB= vapour pressure of pure Benzene =753 mm Hg
Po1= vapour pressure of pure Tolvene =290 mm Hg
xB= mole frection of benzene above the liquid =0.30
Now, potential pressure of benzene above the liquid
PB=xB×PTotal
0.30=PBPTotal.....(1)
Now, Also, potential pressure of benzene above the liquid using Roonth is law
PB=xB liquid×PoB
Similarly potential pressure of Tohve above the liquid
using Raonlt's law
PT=XT liquid×PoT=(1xB liquid)PoT
PTotal=xB liquid×PoB+(1xB liquid)PoT
=PoT+(PoBPoT)xB liquid
using (1)
0.30=PBPOT(PoBPoT)xB liquid=xB liquidPoBPoT+(PoBPoT)xB liquid
Solving, we get
xB liquid=0.30 PoT0.70 PoB0.30 PoT=0.30×290(0.70×7530.30×290)
=0.142
& xT liquid=0.858
Mole percent benzene =14%
Mole percent Tohene =86%

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