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Q1.16:
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 :

Step 1: Calculating Molar Masses and Moles of Components

To determine the mole fraction of each component, we first calculate the number of moles ($n$) using the molar masses of heptane ($C_7H_{16}$) and octane ($C_8H_{18}$).

  • Molar mass of Heptane ($M_{heptane}$): $(7 \times 12.01) + (16 \times 1.008) = 84.07 + 16.128 = 100.2 \, \text{g/mol}$
  • Molar mass of Octane ($M_{octane}$): $(8 \times 12.01) + (18 \times 1.008) = 96.08 + 18.144 = 114.22 \, \text{g/mol}$

Now, calculate the number of moles in the given sample mass:

  • $n_{heptane} = \frac{26.0 \, \text{g}}{100.2 \, \text{g/mol}} \approx 0.2595 \, \text{mol}$
  • $n_{octane} = \frac{35.0 \, \text{g}}{114.22 \, \text{g/mol}} \approx 0.3064 \, \text{mol}$

Step 2: Calculating Mole Fractions

The mole fraction ($\chi$) is defined as the ratio of moles of a component to the total moles in the solution [Per Definition of Mole Fraction].

$n_{total} = n_{heptane} + n_{octane} = 0.2595 + 0.3064 = 0.5659 \, \text{mol}$

  • $\chi_{heptane} = \frac{n_{heptane}}{n_{total}} = \frac{0.2595}{0.5659} \approx 0.4586$
  • $\chi_{octane} = \frac{n_{octane}}{n_{total}} = \frac{0.3064}{0.5659} \approx 0.5414$

Step 3: Calculating Total Vapour Pressure

For an ideal solution, the total vapour pressure ($P_{total}$) is the sum of the partial vapour pressures of the individual components, determined by Raoult's Law [Per Raoult's Law: $P_i = \chi_i P_i^\circ$].

$P_{total} = P_{heptane} + P_{octane} = (\chi_{heptane} \cdot P_{heptane}^\circ) + (\chi_{octane} \cdot P_{octane}^\circ)$

Substituting the given values ($P_{heptane}^\circ = 105.2 \, \text{kPa}$, $P_{octane}^\circ = 46.8 \, \text{kPa}$):

$P_{total} = (0.4586 \times 105.2 \, \text{kPa}) + (0.5414 \times 46.8 \, \text{kPa})$

$P_{total} = 48.2447 \, \text{kPa} + 25.3375 \, \text{kPa} = 73.5822 \, \text{kPa}$

Mole Fraction Composition Vapour Pressure (kPa) 105.2 46.8 P_total = 73.58 kPa

Final Solution: The total vapour pressure of the mixture at 373 K is 73.58 kPa.


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