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Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a specific temperature. In simpler terms, it describes how much a liquid tends to evaporate. Liquids with high vapor pressures evaporate more easily. For example, at 35°C, pure ethanol has a vapor pressure of 100 torr, meaning it readily evaporates under these conditions.

Raoult's Law helps us understand how the vapor pressure of a component behaves when it is part of a mixture. According to the law, the vapor pressure of a component in a solution (\(P_A\)) is proportional to the vapor pressure of the pure component (\(P^*_A\)) times its mole fraction in the mixture. Raoult's Law is expressed as:

• \(P_A = x_A P^*_A\)

Using this, one can calculate how the presence of other substances influences the vapor pressure of a component in a mixture.

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the moles of one component to the total moles in the mixture. In our example exercise, the mole fraction of ethanol (\(x_A\)) tells us how much of the mixture is ethanol.

To find the mole fraction, use this formula:

• \(x_A = \frac{\text{moles of component A}}{\text{total moles in the mixture}}\)

By rearranging Raoult's Law and inserting known values (vapor pressures), the mole fraction of ethanol was determined to be 0.08. This means ethanol makes up 8% of the moles in the mixture. Understanding mole fraction allows us to see how much of each substance is present compared to the entire mixture.

Mixture formulation is the process of combining substances in precise ratios to achieve desired properties, such as a specific vapor pressure. Here, the goal is to reach an 8 torr vapor pressure for ethanol in a mixture with paraffin.

This involves calculating the amount of each component needed, using:

• Molecular weights to convert mass to moles
• The mole fraction to allocate the right quantities of each substance

Specifically, to find the required mass of ethanol, we first determined the moles using the given mole fraction of 0.08. By converting these moles back to mass using ethanol's molecular weight, it was found that approximately 6.9 kg of ethanol should be combined with 620 kg of paraffin to create the desired mixture. This systematic approach ensures the mixture will behave as intended in applications like maintaining a specific vapor pressure for fuel use.