Question

The vapour pressure of a solven decreased by \(10 \mathrm{~mm}\) of \(\mathrm{Hg}\) when a non volatile solute was added to the solvent The mole fraction of solute in th solution is \(0.2 .\) What would be the mol fraction of solvent if decrease in vapou pressure is \(20 \mathrm{~mm}\) of \(\mathrm{Hg}\) ? (a) \(0.8\) (b) \(0.6\) (c) \(0.4\) (d) \(0.7\)

Short answer

The mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg would be 0.6.

Step-by-step solution

Understand Raoult's Law and its implication

According to Raoult's law, the vapor pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent in the solution. Mathematically, it can be expressed as: \( P = P^0_{solvent} \times X_{solvent} \), where \( P \) is the vapor pressure of the solvent when the solute is added, \( P^0_{solvent} \) is the vapor pressure of the pure solvent, and \( X_{solvent} \) is the mole fraction of the solvent. The decrease in vapor pressure is given by \( P^0_{solvent} - P \).

Calculate the initial vapor pressure of the solvent

Let the initial vapor pressure of the pure solvent be \( P^0_{solvent} \). With the first decrease of 10 mm of Hg due to the addition of the solute, the mole fraction of solute is given as 0.2, thus the mole fraction of solvent is 0.8. Let's calculate the new vapor pressure of the solvent after the addition of the solute which we will call \( P_1 \), using the relation \( P_1 = P^0_{solvent} \times 0.8 \).

Relate the decreased vapor pressure to mole fraction

Since the vapor pressure decreased by 10 mm of Hg when the solute was added, we have \( P^0_{solvent} - P_1 = 10 \mathrm{~mm~Hg} \). Substituting the expression for \( P_1 \) from the previous step, the equation becomes \( P^0_{solvent} - P^0_{solvent} \times 0.8 = 10 \), leading to \( P^0_{solvent} \times (1 - 0.8) = 10 \) which implies \( P^0_{solvent} = \frac{10}{0.2} = 50 \mathrm{~mm~Hg} \).

Determine the new mole fraction of solvent for a 20 mm decrease

If the vapor pressure decreases by 20 mm of Hg, let \( P_2 \) be the new vapor pressure and \( X_{solvent2} \) be the new mole fraction of the solvent. According to the problem, \( P^0_{solvent} - P_2 = 20 \mathrm{~mm~Hg} \). With \( P^0_{solvent} = 50 \mathrm{~mm~Hg} \) as calculated, we get \( 50 - P_2 = 20 \) and thus \( P_2 = 30 \mathrm{~mm~Hg} \). Using the relation from Raoult's law, we find \( X_{solvent2} = \frac{P_2}{P^0_{solvent}} = \frac{30}{50} = 0.6 \).

Key concepts

Vapor Pressure

Vapor pressure is a critical property of liquids that describes the pressure exerted by a vapor when it's in equilibrium with its liquid or solid form. This equilibrium occurs when the rate of evaporation of the liquid (or sublimation of the solid) equals the rate of condensation of the vapor.

For pure substances, vapor pressure is a fixed value at a given temperature and is specific to each substance. However, when a non-volatile solute is added to a solvent to create a solution, the vapor pressure of the solvent decreases. This is because the solute particles take up space at the liquid surface, preventing some solvent molecules from vaporizing, leading to a reduction in vapor pressure. This change in vapor pressure can be quantified using Raoult's Law, which provides a relationship between the vapor pressure of the solvent in the solution and the mole fraction of the solvent.

The decrease in vapor pressure is typically measured in units such as millimeters of mercury (mm Hg). In our exercise, the vapor pressure of a solvent decreased by 10 mm Hg upon adding a solute. Understanding this concept is crucial because it underpins the calculations for determining mole fractions and understanding the behavior of solutions.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture or solution. It's defined as the ratio of the number of moles of a particular component to the total number of moles of all components in the mixture. Mathematically, it is denoted as:

\( X_i = \frac{n_i}{n_{total}} \)
where \( X_i \) is the mole fraction of component 'i', \( n_i \) represents the number of moles of 'i', and \( n_{total} \) is the sum of moles of all components.

In the context of our exercise, the mole fraction of solute is given as 0.2, which means that for every 5 moles of solution, there is 1 mole of solute. If we query the same situation but with a 20 mm Hg decrease in vapor pressure, the calculations reveal a mole fraction of 0.6 for the solvent. It's vital for students to grasp this concept, as mole fraction plays a significant role in determining the properties of solutions, including vapor pressure as determined by Raoult's Law.

Ideal Solution

An ideal solution is a special type of solution that adheres to specific criteria, mainly that it follows Raoult's Law throughout the range of possible concentrations. In an ideal solution, the interactions between molecules of different components are similar to those found among molecules of the same component.

For ideal solutions, the vapor pressure of a solvent above a solution is proportional to the mole fraction of the solvent in the solution, and the total vapor pressure can be obtained by summing the partial vapor pressures of each component in the mixture. Raoult's Law can be summarized by the equation:
\( P = P^0_{solvent} \times X_{solvent} \)
where \( P \) is the vapor pressure of the solution, \( P^0_{solvent} \) is the vapor pressure of the pure solvent, and \( X_{solvent} \) is the mole fraction of the solvent.

It's important to acknowledge that not all solutions are ideal. Deviations from this behavior occur when the forces between dissimilar molecules are significantly different from the forces between identical molecules. However, the assumption of ideality is useful for many applications, including the problem in our exercise where the behavior of the solvent upon the addition of a non-volatile solute can be effectively predicted using Raoult's Law.