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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. - CBSE (Science) Class 12 - Chemistry

#### 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?

#### Solution

Vapour pressure of heptane (p_1^0) = 105.2 kPa

Vapour pressure of heptane (p_2^0) = 46.8 kPa

We know that

Molar mass of heptane (C7H16)  = 7 x 12 + 16 x 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, x_1 = 0.26/(0.26 + 0.31)

= 0.456

And, mole fraction of octane, x2 = 1 − 0.456

= 0.544

Now, partial pressure of heptane, p_1=x_1p_1^0

= 0.456 × 105.2

= 47.97 kPa

Partial pressure of octane p_2 = x_2p_2^0

= 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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Solution 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. Concept: Ideal and Non-ideal Solutions.
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