Problem 168 - A Complete Resource Book in Chemistry for JEE Main

Question

On mixing, heptane and octane form an ideal solution. At \(373 \mathrm{~K}\), the vapour pressures of the two liquid components (heptane and octane) are \(105 \mathrm{kPa}\) and \(45 \mathrm{kPa}\) respectively. Vapour pressure of the solution obtained by mixing \(25.0 \mathrm{~g}\) of heptane and \(35 \mathrm{~g}\) of octane will be (molar mass of heptane \(=100 \mathrm{~g} \mathrm{~mol}^{-1}\) and of octane \(=\) \(\left.114 \mathrm{~g} \mathrm{~mol}^{-1}\right)\) (a) \(72.0 \mathrm{kPa}\) (b) \(36.1 \mathrm{kPa}\) (c) \(96.2 \mathrm{kPa}\) (d) \(144.5 \mathrm{kPa}\)

Step-by-Step Solution

Answer

The vapour pressure of the solution is approximately \(72.0 \mathrm{kPa}\) (option a).

Step 1: Calculate Moles of Heptane

First, find the moles of heptane using its molecular weight. \[ n_{ ext{heptane}} = \frac{25.0 \, \text{g}}{100 \, \text{g/mol}} = 0.25 \, \text{mol} \]

Step 2: Calculate Moles of Octane

Next, calculate the moles of octane using its molecular weight.\[ n_{ ext{octane}} = \frac{35.0 \, \text{g}}{114 \, \text{g/mol}} \approx 0.307 \, \text{mol} \]

Step 3: Calculate Mole Fraction of Heptane

Find the mole fraction of heptane using the formula:\[ x_{ ext{heptane}} = \frac{n_{ ext{heptane}}}{n_{ ext{heptane}} + n_{ ext{octane}}} = \frac{0.25}{0.25 + 0.307} \approx 0.449 \]

Step 4: Calculate Mole Fraction of Octane

Determine the mole fraction of octane:\[ x_{ ext{octane}} = \frac{n_{ ext{octane}}}{n_{ ext{heptane}} + n_{ ext{octane}}} = \frac{0.307}{0.25 + 0.307} \approx 0.551 \]

Step 5: Apply Raoult's Law for Partial Pressures

Use Raoult's Law to calculate the partial pressures:\[ P_{ ext{heptane}} = x_{ ext{heptane}} \times P^0_{ ext{heptane}} = 0.449 \times 105 \, \text{kPa} \approx 47.145 \, \text{kPa} \] \[ P_{ ext{octane}} = x_{ ext{octane}} \times P^0_{ ext{octane}} = 0.551 \times 45 \, \text{kPa} \approx 24.795 \, \text{kPa} \]

Step 6: Calculate Total Vapour Pressure

Add the partial pressures to find the total vapour pressure of the solution:\[ P_{ ext{total}} = P_{ ext{heptane}} + P_{ ext{octane}} = 47.145 \, \text{kPa} + 24.795 \, \text{kPa} = 71.94 \, \text{kPa} \]

Key Concepts

Raoult's Law and Its Application

Raoult's Law is a fundamental principle used in chemistry to determine the vapor pressure of a solution. It applies to ideal solutions, which are mixtures where the interactions between different types of molecules are similar to the interactions between the same types of molecules. This means that the presence of one component does not significantly alter the properties, like vapor pressure, of the other.According to Raoult's Law, the partial vapor pressure of a component in a solution (\(P_i\) ) is directly proportional to the mole fraction of that component in the liquid phase (\(x_i\) ) and the vapor pressure of the pure component (\(P_i^0\) ). The equation is given by:\[P_i = x_i \times P_i^0\]To find the total vapor pressure of the solution, you simply add up the partial pressures of all components. This methodology makes Raoult's Law incredibly useful for predicting the properties of mixtures. It gives us insights into how the composition of a solution can alter its overall vapor pressure. However, it's crucial to remember that Raoult's Law only perfectly applies to ideal solutions or those very close to ideal.

Calculating Vapour Pressure of an Ideal Solution

Let's explore how to calculate the vapor pressure of a solution using Raoult's Law, starting with a scenario where heptane and octane are mixed. In this process, we need to consider the individual vapor pressures of the pure components and their respective mole fractions.First, calculate the moles of each component. For heptane, its molecular weight is 100 g/mol, and we have 25.0 g. Hence, the moles of heptane are:\[n_{\text{heptane}} = \frac{25.0}{100} = 0.25 \, \text{mol}\]For octane, with a molecular weight of 114 g/mol and a mass of 35.0 g, the moles of octane are:\[n_{\text{octane}} = \frac{35.0}{114} \approx 0.307 \, \text{mol}\]Next, determine the mole fraction of each component. The mole fraction is the ratio of the moles of one component to the total moles of all components.For heptane:\[x_{\text{heptane}} = \frac{0.25}{0.25 + 0.307} \approx 0.449\]For octane:\[x_{\text{octane}} = \frac{0.307}{0.25 + 0.307} \approx 0.551\]Using these mole fractions, apply Raoult's Law to find the partial pressures:- Heptane: \(P_{ ext{heptane}} = x_{\text{heptane}} \times P^0_{\text{heptane}} = 0.449 \times 105 \, \text{kPa} \approx 47.145 \, \text{kPa}\)- Octane: \(P_{ ext{octane}} = x_{\text{octane}} \times P^0_{\text{octane}} = 0.551 \times 45 \, \text{kPa} \approx 24.795 \, \text{kPa}\)Finally, the total vapor pressure of this ideal solution is the sum of these partial pressures, resulting in about 71.94 kPa.

Understanding Mole Fraction Calculation

The mole fraction is an essential concept in understanding solutions in chemistry, particularly when dealing with vapor pressure calculations. It provides a way to express the concentration of a component in a mixture, vital for applying Raoult's Law.A mole fraction, represented by \(x_i\), is calculated as the ratio of the number of moles of a component to the total number of moles in the mixture. It is a dimensionless quantity and is always less than or equal to one. The sum of the mole fractions of all components in a solution equals one.Here's how to calculate the mole fraction:

• First, determine the moles of each component in the solution.
• Then, add these moles together to get the total moles in the solution.
• Finally, divide the moles of the component of interest by this total.

For example, in a mixture of heptane and octane:

• Moles of heptane: 0.25 mol
• Moles of octane: 0.307 mol
• Total moles = 0.25 + 0.307 = 0.557 mol

Now, the mole fraction of heptane is:\[x_{\text{heptane}} = \frac{0.25}{0.557} \approx 0.449\]And for octane, it is:\[x_{\text{octane}} = \frac{0.307}{0.557} \approx 0.551\]Understanding mole fractions ensures accurate calculations of other properties like vapor pressure and helps explain how mixtures behave under different conditions.