Question

Vapour pressure of $C_6{H}_6$ and $C_7{H}_8$ mixture at $50^{\circ }C$ are given by :
$P_{total}=179X_B+92$ ($X_B=$ mole fraction of $C_6{H}_6$)
Then the vapour pressure of pure benzene in mm of Hg is

• A. 271
• B. 92
• C. 380
• D. 120

Answer 1

Answer: (A)

$P_{total}=179x_B+92$
For a binary volatile liquid mixture,
$P_{total}=p_B^{\circ }{x}_B+p_T^{\circ }{x}_T$
$P_{total}=p_B^{\circ }{x}_B+p_T^{\circ }\left(1-x_B\right)=p_B^{\circ }{x}_B+p_T^{\circ }-p_T^{\circ }{x}_B$
$179x_B+92=\left(p_B^{\circ }-p_T^{\circ }\right)x_B+p_T^{\circ }$
Thus $p_T^{\circ }=92$
$p_B^{\circ }-p_T^{\circ }=179\Rightarrow p_B^{\circ }=179+92=271$ mm of Hg