Question

Vapour pressure of $C_6H_6$ and $C_7H_8$ mixture at $50^{\circ}C$ are given by :
$P_{total} = 179 X_B + 92$ ($X_B =$ mole fraction of $C_6H_6$)
Then the vapour pressure of pure benzene in mm of Hg is

• A. 271
• B. 92
• C. 380
• D. 120

Answer 1

Answer: (A)

$P_{total } = 179 \,x_{B} + 92$
For a binary volatile liquid mixture,
$P_{total} = p^{\circ}_{B} \,x_{B} + p^{\circ}_{T} \,x_{T}$
$P_{total} = p^{\circ}_{B} \,x_{B} + p^{\circ}_{T} \left(1 - x_{B}\right) = p^{\circ}_{B} \,x_{B} + p^{\circ}_{T }- p^{\circ}_{T} \,x_{B}$
$179x_{B} + 92 = \left(p^{\circ}_{B} - p^{\circ}_{T} \right)x_{B} + p^{\circ}_{T}$
Thus $p^{\circ}_{T} = 92$
$p^{\circ}_{B} - p^{\circ}_{T} = 179 \,⇒\, p^{\circ}_{B} = 179 + 92 = 271$ mm of Hg