Problems With Solution - Vapour Pressure Of Liquid-Liquid Solutions

Vapour Pressure of Liquid-Liquid Solutions

In a liquid-liquid solution, the vapour pressure is influenced by the presence of solute molecules in the solvent.
The vapour pressure of a liquid-liquid solution is lower than the vapour pressure of the pure solvent.
This phenomenon is known as the Raoult’s law.

Raoult’s Law

According to Raoult’s law, the partial pressure of each component in the vapour phase of an ideal solution is directly proportional to its mole fraction in the liquid phase.
Mathematically, for a binary liquid-liquid solution, the Raoult’s law equation is given as: $$ P_A = X_A \cdot P^_A \hspace{10mm} P_B = X_B \cdot P^_B $$ Where:
$ P_A $ and $ P_B $ are the partial pressures of components A and B in the vapour phase.
$ X_A $ and $ X_B $ are the mole fractions of components A and B in the liquid phase.
$ P^_A $ and $ P^_B $ are the vapour pressures of pure components A and B.

Positive Deviation from Raoult’s Law

In some cases, the experimental vapour pressure of a liquid-liquid solution is higher than predicted by Raoult’s law.
This is known as positive deviation from Raoult’s law.
Positive deviation occurs when the intermolecular forces between the components of the solution are weaker than those between the similar components (solute-solute and solvent-solvent) in the pure state.

Negative Deviation from Raoult’s Law

In other cases, the experimental vapour pressure of a liquid-liquid solution is lower than predicted by Raoult’s law.
This is known as negative deviation from Raoult’s law.
Negative deviation occurs when the intermolecular forces between the components of the solution are stronger than those between the similar components (solute-solute and solvent-solvent) in the pure state.

Ideal Solutions

An ideal solution is one that follows Raoult’s law exactly.
In an ideal solution, the intermolecular forces between the solute and solvent are similar to those between the solvent-solvent and solute-solute.
The vapour pressure of an ideal solution can be calculated using the Raoult’s law equation.

Colligative Properties

Colligative properties are the properties of solutions that depend only on the number of solute particles and not on their nature.
Here are some examples of colligative properties:
- Vapor pressure lowering
- Boiling point elevation
- Freezing point depression
- Osmotic pressure

Vapor Pressure Lowering

Vapor pressure lowering is the decrease in the vapor pressure of a solvent due to the presence of a non-volatile solute.
It is directly proportional to the mole fraction of solute particles in the solution.
Mathematically, vapor pressure lowering can be calculated using the equation: $$ \Delta P = P^*_0 \cdot X_B $$ Where:
$ \Delta P $ is the vapor pressure lowering
$ P^*_0 $ is the vapor pressure of the pure solvent
$ X_B $ is the mole fraction of the solute

Boiling Point Elevation

Boiling point elevation is the increase in the boiling point of a solvent due to the presence of a non-volatile solute.
It is directly proportional to the molality of the solution and the boiling point elevation constant of the solvent.
Mathematically, boiling point elevation can be calculated using the equation: $$ \Delta T_b = K_b \cdot m $$ Where:
$ \Delta T_b $ is the boiling point elevation
$ K_b $ is the boiling point elevation constant of the solvent
$ m $ is the molality of the solution

Freezing Point Depression

Freezing point depression is the decrease in the freezing point of a solvent due to the presence of a non-volatile solute.
It is directly proportional to the molality of the solution and the freezing point depression constant of the solvent.
Mathematically, freezing point depression can be calculated using the equation: $$ \Delta T_f = K_f \cdot m $$ Where:
$ \Delta T_f $ is the freezing point depression
$ K_f $ is the freezing point depression constant of the solvent
$ m $ is the molality of the solution

Osmotic Pressure

Osmotic pressure is the pressure exerted by a solvent to prevent the inward flow of solvent molecules across a semipermeable membrane.
It is directly proportional to the concentration of the solute particles and the temperature.
Mathematically, osmotic pressure can be calculated using the equation: $$ \Pi = c \cdot R \cdot T $$ Where:
$ \Pi $ is the osmotic pressure
$ c $ is the concentration of the solute particles
$ R $ is the ideal gas constant
$ T $ is the temperature

Problems with Solution - Vapour pressure of liquid-liquid solutions

Problem 1: Calculate the vapour pressure of a solution containing 0.5 moles of solute A and 1.5 moles of solvent B if the vapour pressure of pure A is 40 mmHg and pure B is 60 mmHg.
Solution:
- First, we calculate the mole fraction of each component:
  - $ X_A = \frac{0.5}{0.5 + 1.5} = 0.25 $
  - $ X_B = \frac{1.5}{0.5 + 1.5} = 0.75 $
- Then, we use Raoult’s law equation to calculate the partial pressure of each component:
  - $ P_A = 0.25 \times 40 = 10 $ mmHg
  - $ P_B = 0.75 \times 60 = 45 $ mmHg
- Therefore, the vapour pressure of the solution is 10 mmHg (for A) + 45 mmHg (for B) = 55 mmHg.

Problems with Solution - Positive deviation from Raoult’s Law

Problem 2: A solution of ethanol and acetone shows positive deviation from Raoult’s law. Calculate the vapour pressure of a solution containing 0.4 moles of ethanol and 0.6 moles of acetone, given that the vapour pressures of pure ethanol and acetone are 50 mmHg and 100 mmHg, respectively.
Solution:
- First, we calculate the mole fraction of each component:
  - $ X_{\text{ethanol}} = \frac{0.4}{0.4 + 0.6} = 0.4 $
  - $ X_{\text{acetone}} = \frac{0.6}{0.4 + 0.6} = 0.6 $
- Then, we use Raoult’s law equation to calculate the partial pressure of each component:
  - $ P_{\text{ethanol}} = 0.4 \times 50 = 20 $ mmHg
  - $ P_{\text{acetone}} = 0.6 \times 100 = 60 $ mmHg
- Therefore, the vapour pressure of the solution is 20 mmHg (for ethanol) + 60 mmHg (for acetone) = 80 mmHg.

Problems with Solution - Negative deviation from Raoult’s Law

Problem 3: A solution of chloroform and acetone shows negative deviation from Raoult’s law. Calculate the vapour pressure of a solution containing 0.6 moles of chloroform and 0.4 moles of acetone, given that the vapour pressures of pure chloroform and acetone are 200 mmHg and 100 mmHg, respectively.
Solution:
- First, we calculate the mole fraction of each component:
  - $ X_{\text{chloroform}} = \frac{0.6}{0.6 + 0.4} = 0.6 $
  - $ X_{\text{acetone}} = \frac{0.4}{0.6 + 0.4} = 0.4 $
- Then, we use Raoult’s law equation to calculate the partial pressure of each component:
  - $ P_{\text{chloroform}} = 0.6 \times 200 = 120 $ mmHg
  - $ P_{\text{acetone}} = 0.4 \times 100 = 40 $ mmHg
- Therefore, the vapour pressure of the solution is 120 mmHg (for chloroform) + 40 mmHg (for acetone) = 160 mmHg.

Ideal Solutions - Example

Example 1: A solution containing 2 moles of solute A and 3 moles of solvent B follows Raoult’s law. Calculate the mole fractions of both components and the total vapour pressure of the solution if the vapour pressures of pure A and B are 50 mmHg and 80 mmHg, respectively.
- Solution:
  - $ X_A = \frac{2}{2 + 3} = 0.4 $
  - $ X_B = \frac{3}{2 + 3} = 0.6 $
  - $ P_{\text{total}} = X_A \times P_{\text{A}} + X_B \times P_{\text{B}} = 0.4 \times 50 + 0.6 \times 80 = 60 + 48 = 108 $ mmHg

Colligative Properties - Example

Example 2: Calculate the vapor pressure lowering when 0.5 moles of non-volatile solute glucose is dissolved in 500 mL of water at 25°C. Given that the vapor pressure of pure water at 25°C is 23.8 mmHg.
- Solution:
  - Calculate the mole fraction of glucose:
    - $ X_{\text{glucose}} = \frac{0.5}{0.5 + 55.55} = 0.0089 $
  - Calculate the vapor pressure lowering using the equation $ \Delta P = P^*0 \times X{\text{glucose}} $ :
    - $ \Delta P = 23.8 \times 0.0089 = 0.212 $ mmHg

Colligative Properties - Boiling Point Elevation

Boiling point elevation is the increase in the boiling point of a solvent when a non-volatile solute is added.
The boiling point elevation is given by the equation $ \Delta T_b = K_b \times m $ , where $ K_b $ is the boiling point elevation constant and $ m $ is the molality of the solution.

Colligative Properties - Freezing Point Depression

Freezing point depression is the decrease in the freezing point of a solvent when a non-volatile solute is added.
The freezing point depression is given by the equation $ \Delta T_f = K_f \times m $ , where $ K_f $ is the freezing point depression constant and $ m $ is the molality of the solution.

Colligative Properties - Osmotic Pressure

Osmotic pressure is the pressure exerted by a solvent to prevent the inward flow of solvent molecules across a semipermeable membrane.
Osmotic pressure is given by the equation $ \Pi = c \times R \times T $ , where $ c $ is the concentration of the solute particles, $ R $ is the ideal gas constant, and $ T $ is the temperature.

Colligative Properties - Application

A practical application of colligative properties is in the determination of molecular weights of substances.
By measuring the osmotic pressure of a solution, the molecular weight of the solute can be calculated using the equation $ \Pi = \frac{n}{V} \times R \times T $ .
This method is known as osmometry.

Summary

Vapour pressure of liquid-liquid solutions is influenced by the presence of solute molecules in the solvent.
Raoult’s law describes the relationship between the partial pressure of each component and its mole fraction in the liquid phase.
Positive deviation from Raoult’s law occurs when the intermolecular forces between the components are weaker than those in the pure state.
Negative deviation occurs when the intermolecular forces between the components are stronger than those in the pure state.
Colligative properties are properties of solutions that depend on the number of solute particles and not their nature.
Examples of colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.

Vapor Pressure of Liquid-Liquid Solutions

Vapor pressure is the pressure exerted by the vapor of a substance when it is in equilibrium with its liquid phase.
In a liquid-liquid solution, the vapor pressure is influenced by the presence of solute molecules in the solvent.
The vapor pressure of the solution is lower than the vapor pressure of the pure solvent due to the solute-solvent interactions.

Raoult’s Law

Raoult’s law describes the relationship between the vapor pressure of a component in a liquid-liquid solution and its mole fraction in the liquid phase.
According to Raoult’s law, the partial pressure of each component in the vapor phase is directly proportional to its mole fraction in the liquid phase.
Mathematically, for a binary liquid-liquid solution, Raoult’s law can be expressed as:
- $ P_A = X_A \cdot P_A^* $
- $ P_B = X_B \cdot P_B^* $ Where:
$ P_A $ and $ P_B $ are the partial pressures of components A and B in the vapor phase.
$ X_A $ and $ X_B $ are the mole fractions of components A and B in the liquid phase.
$ P_A^* $ and $ P_B^* $ are the vapor pressures of pure components A and B.

Positive Deviation from Raoult’s Law

In some cases, the experimental vapor pressure of a liquid-liquid solution is higher than predicted by Raoult’s law.
This occurrence is known as positive deviation from Raoult’s law.
Positive deviation arises when the solute-solute and solvent-solvent interactions are stronger than solute-solvent interactions.

Negative Deviation from Raoult’s Law

In other cases, the experimental vapor pressure of a liquid-liquid solution is lower than predicted by Raoult’s law.
This phenomenon is known as negative deviation from Raoult’s law.
Negative deviation arises when the solute-solute and solvent-solvent interactions are weaker than solute-solvent interactions.

Ideal Solutions

An ideal solution is one that follows Raoult’s law exactly.
In an ideal solution, the intermolecular forces between the solute and solvent are similar to those between the solvent-solvent and solute-solute.
The vapor pressure of an ideal solution can be calculated using Raoult’s law equation.

Colligative Properties

Colligative properties are the properties of solutions that depend only on the number of solute particles and not on their nature.
These properties include:
- Vapor pressure lowering
- Boiling point elevation
- Freezing point depression
- Osmotic pressure

Vapor Pressure Lowering

Vapor pressure lowering is the decrease in the vapor pressure of a solvent due to the presence of a non-volatile solute.
It is directly proportional to the mole fraction of the solute particles in the solution.
Mathematically, vapor pressure lowering can be calculated using the equation:
- $ \Delta P = P_A^* \cdot X_B $ Where:
$ \Delta P $ is the vapor pressure lowering
$ P_A^* $ is the vapor pressure of the pure solvent
$ X_B $ is the mole fraction of the solute

Boiling Point Elevation

Boiling point elevation is the increase in the boiling point of a solvent due to the presence of a non-volatile solute.
It is directly proportional to the molality of the solution and the boiling point elevation constant of the solvent.
Mathematically, boiling point elevation can be calculated using the equation:
- $ \Delta T_b = K_b \cdot m $ Where:
$ \Delta T_b $ is the boiling point elevation
$ K_b $ is the boiling point elevation constant of the solvent
$ m $ is the molality of the solution

Freezing Point Depression

Freezing point depression is the decrease in the freezing point of a solvent due to the presence of a non-volatile solute.
It is directly proportional to the molality of the solution and the freezing point depression constant of the solvent.
Mathematically, freezing point depression can be calculated using the equation:
- $ \Delta T_f = K_f \cdot m $ Where:
$ \Delta T_f $ is the freezing point depression
$ K_f $ is the freezing point depression constant of the solvent
$ m $ is the molality of the solution

Osmotic Pressure

Osmotic pressure is the pressure exert