Heptane and octane form ideal solution. At \(373 \mathrm{~K}\), the vapour pressures of the pure liquids are \(106 \mathrm{kPa}\) and \(46 \mathrm{kPa}\), respectively. What will be the vapour pressure, in bar, of a mixture of \(30.0 \mathrm{~g}\) of heptane and \(34.2 \mathrm{~g}\) of octane? (a) 76 bar (b) 152 bar (c) \(1.52\) bar (d) \(0.76\) bar

An ideal solution is a mixture wherein the intermolecular forces between different components are similar to the forces among like molecules. In more tangible terms, this means that when two substances are mixed to form an ideal solution, their interactions with each other are so seamless that the solution's properties are linear combinations of the properties of each pure substance.

In our exercise, heptane and octane combine to form such an ideal solution. These components don't react with each other; they mix thoroughly without generating heat or causing volume changes. This concept is crucial because it sets the stage for applying Raoult's Law, which simplifies the calculation of vapour pressures in such solutions. For an ideal solution, Raoult's Law can predict the resulting vapour pressure based on the proportion of each component present.

Vapour pressure is a critical concept in thermodynamics and physical chemistry. It's the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. The ability to calculate vapour pressure has practical implications in various scientific fields, including meteorology, cooking, and the development of substances in industrial chemistry.

In the context of the exercise, Raoult's Law states that the vapour pressure of an ideal solution is equal to the sum of the partial pressures of each component. The partial pressure is found by multiplying the mole fraction of a component by its pure vapour pressure. A proper understanding of this relation is important not only for solving textbook problems but also for predicting the physical behavior of mixtures in real-life applications.

The mole fraction represents the ratio of moles of a component to the total moles in the mixture. It is a unitless quantity which provides a way to express the concentration of a component within a mixture. In our steps to solution, calculating the mole fractions is an essential intermediate step before we can apply Raoult's Law.

Understanding how mole fraction impacts vapour pressure is key to mastering this concept. In the exercise, once the masses of heptane and octane were converted to moles, their respective mole fractions were calculated. These values are critical because they tell us about the relative amount of each component in the vapor phase. Especially for students in chemistry, becoming comfortable with converting between mass, moles, and mole fractions is a fundamental skill that underpins many aspects of the subject.