Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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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. - 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

Is there an error in this question or solution?