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  • Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Heptane and octane form an ideal solutio | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Q16.

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Vapour pressure of heptane p10 = 105.2 kPa

Vapour pressure of octane p20= 46.8 kPa

As we know that, Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1 = 100 g mol - 1

∴ Number of moles of heptane = 26/100 mol  = 0.26 mol

Molar mass of octane (C8H18) = 8 × 12 + 18 × 1 = 114 g mol - 1

∴ Number of moles of octane = 35/114 mol = 0.31 mol

Mole fraction of heptane, x1 = 0.26 / 0.26 + 0.31

= 0.456

And, mole fraction of octane, x2 = 1 - 0.456 = 0.544

Now, partial pressure of heptane, p1 = x2 p20

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane,p2 = x2 p20

= 0.544 × 46.8 = 25.46 kPa

Hence, vapour pressure of solution, ptotal  =  p1 + p2

= 47.97 + 25.46

= 73.43 kPa

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What is the correct answer to: Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane??

Vapour pressure of heptane p10 = 105.2 kPa

Vapour pressure of octane p20= 46.8 kPa

As we know that, Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1 = 100 g mol - 1

∴ Number of moles of heptan...

How do you solve Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane? step by step?

Step-by-step explanation:
• Vapour pressure of heptane p10 = 105
• 2 kPa
•
• Vapour pressure of octane p20= 46
• 8 kPa

What common mistakes should be avoided in this question?

Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

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Student Discussion

NATS
Class · · , · Oct 04, 2023
35/114=0.30 .....but why 0.31?
👍 👎 Reply
Nayan M J
Class · · , · Jun 26, 2019
It was a suitable answer
👍 👎 Reply
Fucker
Class · · , · May 15, 2019
There is a mistake for finding partial pressure. It should have been written Pa = xap°a
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Aakarshan
Class · · , · Apr 24, 2019
The yellow background soothes me.
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chutiya
Class · · , · Apr 22, 2019
dhanywad
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Procrastinator
Class · · , · Mar 12, 2019
God bless ur soul
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Shivam Yadav
Class · · , · Mar 09, 2019
Wow thanks
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Alfin
Class · · , · Mar 01, 2019
p1 = x2 p20??
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Samoon karamat
Class · · , · May 30, 2018
Thanks for this help ....best &simple solution
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Rama
Class · · , · Oct 17, 2017
Thanks for this help
Best site!!!!!!
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