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On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components(heptane and octane) are 105 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing 25.0 gof heptane and 35 g of octane will be(molar mass of heptane = 100 g mol–1 and of octane = 114 gmol–1) [2010]
  • a)
    72.0 kPa
  • b)
    36.1 kPa
  • c)
    96.2 kPa
  • d)
    144.5 kPa
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
On mixing, heptane and octane form an ideal solution. At 373 K, the va...
Using Raoult's law:-Total vapour pressure=(partial vapour pressure of heptane + partial vapour pressure of octane)= P1X1 + P2X2Where X is mole fraction of component
100 g heptane = 1 mole. 25 g heptane = 0.25 moles. 114 g octane = 1 mole35 g octane = 35/114 = 0.307 moles
Mole fraction of heptane = n1/(n1 + n2)=0.25/(0.25+0.307)=0.4488
Mole fraction of octane = n2/(n1 + n2)=0.307/(0.25+0.307)=0.5511
Applying this in Raoult's law105(0.4488)+45(0.5511)You get 72 kPa
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On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components(heptane and octane) are 105 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing 25.0 gof heptane and 35 g of octane will be(molar mass of heptane = 100 g mol–1 and of octane = 114 gmol–1) [2010]a)72.0 kPab)36.1 kPac)96.2 kPad)144.5 kPaCorrect answer is option 'A'. Can you explain this answer?
Question Description
On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components(heptane and octane) are 105 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing 25.0 gof heptane and 35 g of octane will be(molar mass of heptane = 100 g mol–1 and of octane = 114 gmol–1) [2010]a)72.0 kPab)36.1 kPac)96.2 kPad)144.5 kPaCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components(heptane and octane) are 105 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing 25.0 gof heptane and 35 g of octane will be(molar mass of heptane = 100 g mol–1 and of octane = 114 gmol–1) [2010]a)72.0 kPab)36.1 kPac)96.2 kPad)144.5 kPaCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components(heptane and octane) are 105 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing 25.0 gof heptane and 35 g of octane will be(molar mass of heptane = 100 g mol–1 and of octane = 114 gmol–1) [2010]a)72.0 kPab)36.1 kPac)96.2 kPad)144.5 kPaCorrect answer is option 'A'. Can you explain this answer?.
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