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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 of Components

First, we determine the molar mass of Heptane ($C_7H_{16}$) and Octane ($C_8H_{18}$):

  • Molar mass of Heptane ($M_A$) = $(7 \times 12.01) + (16 \times 1.008) = 84.07 + 16.128 = 100.20 \text{ g/mol}$
  • Molar mass of Octane ($M_B$) = $(8 \times 12.01) + (18 \times 1.008) = 96.08 + 18.144 = 114.22 \text{ g/mol}$

Step 2: Calculating Number of Moles of Each Component

Given the masses $w_A = 26.0 \text{ g}$ and $w_B = 35.0 \text{ g}$, we apply the formula $n = \frac{w}{M}$:

  • $n_{\text{Heptane}} = \frac{26.0}{100.20} \approx 0.2595 \text{ mol}$
  • $n_{\text{Octane}} = \frac{35.0}{114.22} \approx 0.3064 \text{ mol}$
  • Total moles ($n_{total}$) = $0.2595 + 0.3064 = 0.5659 \text{ mol}$

Step 3: Calculating Mole Fractions ($x$)

Using the definition of mole fraction $x_i = \frac{n_i}{n_{total}}$ [Per Dalton's Law concepts]:

  • $x_{\text{Heptane}} = \frac{0.2595}{0.5659} \approx 0.4586$
  • $x_{\text{Octane}} = \frac{0.3064}{0.5659} \approx 0.5414$

Step 4: Applying Raoult's Law for Ideal Solutions

The total vapour pressure $P_{total}$ of an ideal solution is given by: $P_{total} = P_A^\circ x_A + P_B^\circ x_B$ [Per Raoult's Law].

Component Pure Vapour Pressure ($P^\circ$) Mole Fraction ($x$)
Heptane 105.2 kPa 0.4586
Octane 46.8 kPa 0.5414

Step 5: Final Calculation

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

$P_{total} = 48.2447 + 25.3375$

$P_{total} = 73.5822 \text{ kPa}$

Visual Representation of Ideal Vapour Pressure Mixing

P (kPa) 0 1 Mole Fraction of Octane P_total

Final Solution: The vapour pressure of the mixture is 73.58 kPa.


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