The vapour pressure of a benzene – toluene binary solution at certain temperature is given by $P = 450 - 100 X_{T}$ .
Where $X_{T}$ is the mole fraction of toluene. Find the vapour pressures of Benzene and Toluene in pure state.

P_{B}^{o} = 350 P_{T}^{o} = 450

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P_{B}^{o} = 150 P_{T}^{o} = 350

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P_{B}^{o} = 450 P_{T}^{o} = 350

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P_{B}^{o} = 350 P_{T}^{o} = 150

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Solution

The correct option is C $P_{B}^{o} = 450 P_{T}^{o} = 350$
$P_{T}^{o}$ and $P_{B}^{o}$ = vapour pressure of pure toluene and pure benzene respectively.
$P = P_{B}^{o} X_{B} + P_{T}^{o} X_{T}$
$= P_{B}^{o} (1 - X_{T}) + P_{T}^{o} X_{T}$
$= P_{B}^{o} - P_{B}^{o} X_{T} + P_{T}^{o} X_{T}$
$= P_{B}^{o} + (P_{T}^{o} - P_{B}^{o}) X_{T} - - - - - (1)$
$P = 450 - 100 X - - - - - (2)$
Comparing both equations
$= P_{B}^{o} = 450$
$= P_{T}^{o} - 450 = - 100$
$P_{T}^{o} = 350$

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The vapour pressure of a benzene – toluene binary solution at certain temperature is given by $P = 450 - 100 X_{T}$ .
Where $X_{T}$ is the mole fraction of toluene. Find the vapour pressures of Benzene and Toluene in pure state.

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The vapour pressure of a benzene – toluene binary solution at certain temperature is given by P=450–100XT. Where XT is the mole fraction of toluene. Find the vapour pressures of Benzene and Toluene in pure state.

The correct option is C PoB=450 PoT=350PoT and PoB = vapour pressure of pure toluene and pure benzene respectively. P=PoB XB+PoT XT =PoB(1−XT)+PoT XT =PoB−PoBXT+PoT XT =PoB+(PoT−PoB)XT−−−−−(1) P=450–100X−−−−−(2) Comparing both equations =PoB=450 =PoT−450=−100 PoT=350

The vapour pressure of a benzene – toluene binary solution at certain temperature is given by $P = 450 - 100 X_{T}$ .
Where $X_{T}$ is the mole fraction of toluene. Find the vapour pressures of Benzene and Toluene in pure state.