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JEE Main 20204 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2020 · 4 September · Morning Shift · Chemistry · Solutions

At 300K , the vapour pressure of a solution containing 1 mole of n...

At 300K300K, the vapour pressure of a solution containing 11 mole of nn -hexane and 33 moles of nn -heptane is 550mm550mm of HgHg. At the same temperature, if one more mole of nn -heptane is added to this solution, the vapour pressure of the solution increases by 10mm10mm of HgHg. What is the vapour pressure in mmHgmmHg of nn - heptane in its pure state______?

Correct answer

0

Step-by-step solution

Ptotal=P∘hexane.Xhexane+P∘heptane.Xheptane{P}_{total} = {P}^{∘}hexane . {X}_{hexane} + {P}^{∘}heptane . {X}_{heptane}

550=[P∘hexane]×14+[P∘heptane]×34550 = \left[{P}^{∘}hexane\right] \times \dfrac{1}{4} + \left[{P}^{∘}heptane\right] \times \dfrac{3}{4} ..(i)( i )

After mixing 11 mole n-heptane

560=[P∘hexane]×15+[P∘heptane]×45560 = \left[{P}^{∘}hexane\right] \times \dfrac{1}{5} + \left[{P}^{∘}heptane\right] \times \dfrac{4}{5} .....(ii)( ii )

On solving eq. (i) and (ii)

P∘heptane=600mm{P}^{∘}heptane = 600mm of HgHg

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