P _ A =(235 y -125 xy ) mm of Hg where P A is partial pressure of A x is mole fraction of…
Chemistry · JEE Main · NTA Exams — Solutions
\(\mathrm{P}_{\mathrm{A}}=(235 \mathrm{y}-125 \mathrm{xy})\) mm of Hg
where PA is partial pressure of A
x is mole fraction of B in liquid phase in the mixture of two liquids A and B and y is mole fraction of A in vapour phase, then \(\mathrm{F}_{\mathrm{B}}^{\mathrm{a}}\) in mm of Hg is:
- 235
- 0
- 125
- 110
Answer
(D) 110
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