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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 - Chemistry

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प्रश्न

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?

संख्यात्मक
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उत्तर १

Vapour pressure of heptane `(p_1^circ)` = 105.2 kPa

Vapour pressure of octane `(p_2^circ)` = 46.8 kPa

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 = `x_1 p_1^circ`

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane p2 = `x_2 p_2^circ`

= 0.544 × 46.8

= 25.46 kPa

Hence, the vapour pressure of the solution, `p_"total"` = p1 + p2

= 47.97 + 25.46

= 73.43 kPa

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उत्तर २

No. of moles of heptane = `26/100`    ..(Mol. mass of heptane = 100 g mol−1)

= 0.260

 No. of moles of octane = `35/114`    ...(Mol. mass of octane = 114 g mol−1)

= 0.307

∴ Total no. of moles in solution = 0.260 + 0.307

= 0.567

Hence, mole fraction of heptane = `0.260/0.567`

= 0.458

Mole fraction of octane = `0.307/0.567`

= 0.541

Since the solution is ideal,

`p = p_"heptane" + p_"octane"`

= (0.458 × 105.2) + (0.541 × 46.8) kpa

= 73.50 kpa

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पाठ 1: Solutions - Exercises [पृष्ठ २८]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 12
पाठ 1 Solutions
Exercises | Q 1.16 | पृष्ठ २८
नूतन Chemistry Part 1 and 2 [English] Class 12 ISC
पाठ 2 Solutions
'NCERT TEXT-BOOK' Exercises | Q 2.16 | पृष्ठ १२५

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