A) The vapour will contain a higher percentage of benzene
B) The vapour will contain a higher percentage of toluene
C) The vapour will contain equal amounts of benezene and toluene
D) Not enough information is given to make a predication
Correct Answer: A
Solution :
| [a] A\[\to \] benzene, \[\to \] toluene |
| 1 : 1 molar mixture of A and B |
| \[\therefore \] \[{{x}_{A}}=\frac{1}{2}\]and \[{{x}_{B}}=\frac{1}{2}\] |
| \[P{{}_{s}}=P_{A}^{0}{{X}_{A}}+P_{B}^{0}{{X}_{B}}\] |
| \[{{P}_{s}}=12.8\times \frac{1}{2}+3.85\times \frac{1}{2\,}=8.325\,\text{kPa}\] |
| \[{{Y}_{A}}=\frac{P_{A}^{0}{{X}_{A}}}{{{P}_{s}}}=\frac{12.8\times \frac{1}{2}}{8.325}=0.768\] |
| \[\therefore \] \[{{Y}_{B}}=1-{{Y}_{A}}=1-0.768=0.232\] |
| so, the vapour will contain higher percentage of benzene. |
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