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Vapour pressure is a key concept in understanding how substances interact when they are combined in a solution. It is defined as the pressure exerted by the vapor in equilibrium with its liquid or solid form. This equilibrium is established when molecules escape from the liquid phase and enter the vapor phase at a constant rate, much like a fine mist above a calm lake.
In our context, we look at how the vapour pressure of a solvent changes when a solute (usually non-volatile) is added. The addition of a non-volatile solute decreases the vapour pressure of the solvent. This effect can be explained by the fact that solute molecules occupy space at the surface level where molecules escape into the gaseous state.
Raoult's Law serves as a useful tool here. It quantifies this decrease in vapour pressure in direct proportion to the mole fraction of the solute. For example, if a solvent has a vapour pressure of 185 torr, like pure acetone at a given temperature, and decreases to 183 torr upon adding a solute, it means the solute is actively affecting how readily acetone molecules can transition to the gas phase.

The mole fraction is a way of expressing the concentration of a component in a mixture. It's calculated by dividing the number of moles of one component by the total number of moles in the mixture. This value gives us an idea of the proportion of the entire mixture made up by that specific component.
When applying Raoult’s Law, understanding the mole fraction is crucial. It is symbolized as either \(X\), \(X_{solute}\) for solute, or \(X_{solvent}\) for solvent. For instance, in our exercise, the mole fraction is used to determine how the presence of the solute has altered the vapour pressure of acetone.
To find the mole fraction of the solute in our solution, we use the change in vapour pressure as part of Raoult's Law formula: \( \Delta P = P_{solvent} \times X_{solute} \). Solving for the mole fraction gives context to how much of a presence the solute has in the solution.

A non-volatile solute is a substance that does not easily vaporize under existing conditions. Instead of readily escaping into the gas phase, it remains dissolved in the solvent. This property of non-volatility makes it a significant factor in altering the physical properties of solutions.
In our exercise, a non-volatile solute was added to acetone, affecting its vapour pressure. Since the solute does not contribute to vapour pressure, it only impacts the solvent’s behavior. By decreasing the number of molecules that can escape into the gas phase, the non-volatile component significantly lowers the vapour pressure.
The decrease in vapour pressure is helpful when calculating the molarity of a solute in a given solution. For instance, understanding that the solute's presence decreases the vapour pressure from 185 to 183 torr can help us deduce its mole fraction, leading us closer to its molar mass.

Finding the molar mass of a solute involves integrating various pieces of information gathered from the experiment. Molar mass is a measure of the number of grams per mole of a substance, and it's instrumental in identifying and understanding substances.
In the exercise, we understood the change in vapour pressure, determined the mole fraction of the solute, and used these to identify the number of moles present. With the weight of the solute provided (1.2 grams) and the calculated number of moles (approximately 0.0187 moles), we can determine the molar mass using the formula:

• \( Molar \ Mass \ (M) = \frac{mass\ of\ solute}{moles\ of\ solute} \)

Substituting the values into this formula gives us a molar mass of about 64 g/mol, allowing us to select the correct option from the choices given.