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 vapor pressure of the solution is approximately 72.0 kPa, corresponding to option (a).

Step 1: Calculate Moles of Heptane

First, determine the moles of heptane using its mass and molar mass. The formula is \( n = \frac{m}{M} \), where \( n \) is the number of moles, \( m \) is the mass, and \( M \) is the molar mass.\[ n_{ ext{heptane}} = \frac{25.0 \, \text{g}}{100 \, \text{g/mol}} = 0.25 \, \text{mol}\]

Step 2: Calculate Moles of Octane

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

Step 3: Calculate Mole Fractions

Now, find the mole fractions of heptane and octane in the solution. The formula for mole fraction \( X \) is:\[X = \frac{n}{n_{ ext{total}}}\]where \( n_{ ext{total}} = n_{ ext{heptane}} + n_{ ext{octane}} = 0.25 + 0.307 = 0.557 \, \text{mol}\).For heptane:\[X_{ ext{heptane}} = \frac{0.25}{0.557} \approx 0.449\]For octane:\[X_{ ext{octane}} = \frac{0.307}{0.557} \approx 0.551\]

Step 4: Calculate Partial Pressures

Using Raoult's Law, calculate the partial vapor pressures of each component. Raoult's law states that the partial pressure \( P_i \) is the product of the mole fraction and vapor pressure of the pure component:- For heptane:\[P_{ ext{heptane}} = X_{ ext{heptane}} \times P_{ ext{heptane, pure}} = 0.449 \times 105 \, \text{kPa} = 47.145 \, \text{kPa}\]- For octane:\[P_{ ext{octane}} = X_{ ext{octane}} \times P_{ ext{octane, pure}} = 0.551 \times 45 \, \text{kPa} = 24.795 \, \text{kPa}\]

Step 5: Calculate Total Vapor Pressure

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

Step 6: Conclusion

The total vapor pressure of the solution is approximately \(72.0 \, \text{kPa}\). This matches option (a).

Key Concepts

Ideal Solution

In chemistry, an ideal solution is a mixture where the enthalpy of mixing is zero, and the components follow Raoult's Law perfectly. This means that the interactions between the molecules in the mixture are identical to the interactions between the molecules of each pure component. Thus, when heptane and octane are mixed, they form an ideal solution. Their linear hydrocarbons have similar structures and sizes, allowing them to interact more or less similarly with each other. This similarity explains why their mixture behaves ideally.

• No energy is absorbed or released during mixing, indicating no enthalpic interactions such as hydrogen bonding or significant van der Waals forces are at play other than those within the pure components.
• The volume of the solution is simply the sum of the volumes of the individual components because their molecular interactions are unchanged.

This concept simplifies predictions of solutions' behaviors significantly, especially when calculating properties such as vapor pressure.

Vapor Pressure

Vapor pressure is a measure of the tendency of a substance to evaporate. In the context of an ideal solution, each component's vapor pressure contributes to the total vapor pressure of the solution. For example, in our case with heptane and octane, you calculate the vapor pressure of each component using Raoult's Law.
According to Raoult's Law, the partial vapor pressure of each component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution:

• Heptane's pure vapor pressure at 373 K is 105 kPa.
• Octane's pure vapor pressure at 373 K is 45 kPa.

The total vapor pressure is then the sum of these partial pressures. This approach is practical because it links directly to measurable quantities - the vapor pressure of pure substances at a given temperature and their mole fractions.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the number of moles of a component divided by the total number of moles in the mixture. Understanding mole fraction is vital for calculating several properties, including vapor pressure in an ideal solution.
For the heptane and octane mixture in the problem:

• The mole fraction of heptane was calculated by dividing the moles of heptane (0.25 mol) by the total moles of both components (0.557 mol), resulting in approximately 0.449.
• Similarly, the mole fraction of octane was found as the moles of octane (0.307 mol) divided by the total moles (0.557 mol), giving around 0.551.

These mole fractions indicate how much of each component is present in the mixture and are crucial for applying Raoult's Law to predict the properties of the solution accurately.
In essence, mole fraction provides a clear picture of the proportion of each component in a solution and is an essential component in thermodynamic calculations.