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Standard XII

Chemistry

Mole Fraction

Heptane and o...

Question

Heptane and octane form an ideal solution. At $373 K$ , the vapour pressures of the two liquid components are $105.2 k P a$ and $46.8 k P a$ 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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Solution

The molar masses of heptane and octane are $100$ g/mol and $114$ g/mol respectively.

$26$ g of heptane corresponds to $\frac{26}{100} = 0.26$ moles

$35$ g of octane corresponds to $\frac{35}{114} = 0.31$ moles

Mole fraction of heptane $χ = \frac{0.26}{0.26 + 0.31} = 0.456$

Mole fraction of octane $χ^{'} = 1 - 0.456 = 0.544$

Partial pressure of heptane $p = 0.456 \times 105.2 = 47.97$ kPa

Partial pressure of octane $p^{'} = 0.544 \times 46.8 = 25.46$ kPa

Vapour pressure of solution $P = p + p^{'} = 47.97 + 25.46 = 73.43$ kPa

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QUESTION 2.156

Heptane and octane form an ideal solution. At 373 K, the vapour pressure of the two liquid components is 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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Q. On mixing, heptane and octane form an ideal solution. At 373

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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 solution. At $373 K$ , the vapour pressures of the two liquid components are $105.2 k P a$ and $46.8 k P a$ respectively. What will be the vapour pressure of a mixture of $26.0 g$ of heptane and $35 g$ of octane?