© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the formula for total vapour pressure of an ideal solution
According to Raoult’s Law, for an ideal solution containing two volatile components A and B, the total vapour pressure, $P_{\text{Total}}$, is given by:
$P_{\text{Total}} = P_A^\circ \, x_A \;+\; P_B^\circ \, x_B$
Here, $P_A^\circ$ and $P_B^\circ$ are the vapour pressures of the pure components A and B respectively, and $x_A$ and $x_B$ are their respective mole fractions in the solution.
Step 2: Identify components and given data
In this problem:
Heptane (component A) has $P_A^\circ = 105\,\text{kPa}$ and molar mass $= 100\,\text{g mol}^{-1}$.
Octane (component B) has $P_B^\circ = 45\,\text{kPa}$ and molar mass $= 114\,\text{g mol}^{-1}$.
Mass of heptane $= 25.0\,\text{g}$.
Mass of octane $= 35.0\,\text{g}$.
Temperature $= 373\,\text{K}$ (for reference, though the temperature is already used in vapour pressures provided).
Step 3: Calculate the moles of each component
The number of moles of a substance is given by:
$\text{moles} = \frac{\text{mass of substance}}{\text{molar mass}}$
So, moles of heptane ($n_{\text{hep}}$):
$n_{\text{hep}} = \frac{25.0 \,\text{g}}{100 \,\text{g mol}^{-1}} = 0.25\,\text{mol}$
Moles of octane ($n_{\text{oct}}$):
$n_{\text{oct}} = \frac{35.0 \,\text{g}}{114 \,\text{g mol}^{-1}} \approx 0.307\,\text{mol} \,(\text{approx. }0.30\,\text{mol})$
Step 4: Calculate the mole fractions
The total moles of the mixture is:
$n_{\text{total}} = n_{\text{hep}} + n_{\text{oct}} \approx 0.25 + 0.30 = 0.55\,\text{mol}$
Hence, the mole fractions are:
$x_{\text{hep}} = \frac{n_{\text{hep}}}{n_{\text{total}}} = \frac{0.25}{0.55}$
$x_{\text{oct}} = \frac{n_{\text{oct}}}{n_{\text{total}}} = \frac{0.30}{0.55}$
Step 5: Substitute values into Raoult’s Law expression
Using $P_{\text{Total}} = P_A^\circ \, x_A + P_B^\circ \, x_B$:
$P_{\text{Total}} = 105 \times \frac{0.25}{0.55} + 45 \times \frac{0.30}{0.55}$
Calculate each term:
First term: $105 \times \frac{0.25}{0.55} \approx 105 \times 0.4545 \approx 47.7\,\text{kPa}$
Second term: $45 \times \frac{0.30}{0.55} \approx 45 \times 0.5455 \approx 24.55\,\text{kPa}$
Sum of both terms:
$P_{\text{Total}} \approx 47.7 + 24.55 = 72.25 \,\text{kPa} \approx 72.0 \,\text{kPa}$
Step 6: State the final answer
Therefore, the vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35.0 g of octane at 373 K is approximately $72.0\,\text{kPa}$.
