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 1

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

Answer 2

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

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

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

= 48.1816 + 25.3188

= 73.50 kpa