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heptane and octane form an ideal solution at 373...

Question:
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?

Answer:

Vapour pressure of heptane p₁⁰ = 105.2 kPa

Vapour pressure of octane p_²⁰= 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= x₂p₂⁰

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane,p₂= x₂p₂⁰

= 0.544 × 46.8 = 25.46 kPa

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

= 47.97 + 25.46

= 73.43 kPa

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Question:
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?

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Question: 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?

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Question:
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?