Raoult's Law for Ideal Solutions
1. Definition and Core Explanation
Raoult's Law states that in an ideal solution, the partial vapor pressure of each component is directly proportional to its mole fraction in the solution. Mathematically, it is expressed as:
\[
P_A = P_A^\circ \cdot x_A
\]
\[
P_B = P_B^\circ \cdot x_B
\]
where:
- \( P_A \) and \( P_B \) are the partial vapor pressures of components \( A \) and \( B \),
- \( P_A^\circ \) and \( P_B^\circ \) are the vapor pressures of the pure components, and
- \( x_A \) and \( x_B \) are the mole fractions of the components in the solution.
In an ideal solution, the interactions between molecules of different components are similar to those between molecules of the same component. As a result, there is no significant enthalpy change or volume change when mixing the components.
2. Fundamental Properties of Ideal Solutions
- Additive Partial Pressures: The total vapor pressure of an ideal solution is the sum of the partial vapor pressures of its components:
\[
P_{\text{total}} = P_A + P_B = P_A^\circ \cdot x_A + P_B^\circ \cdot x_B
\]
- Mole Fraction Dependency: Each component's vapor pressure contribution depends on its mole fraction, leading to predictable changes as composition varies.
- No Heat or Volume Change: Ideal solutions have no enthalpy or volume change on mixing, as intermolecular forces between different molecules are similar to those within each component.
3. Key Terms and Concepts
- Ideal Solution: A solution that obeys Raoult's Law across all concentrations.
- Partial Vapor Pressure: The pressure exerted by each component in a mixture if it alone occupied the container.
- Vapor Pressure: The pressure exerted by a vapor in equilibrium with its liquid phase.
4. Important Rules, Theorems, and Principles
- Raoult’s Law Validity: Applies to ideal solutions, where molecular interactions in the mixture are similar to those in pure components.
- Non-Ideal Solutions: Deviations from Raoult's Law occur in non-ideal solutions due to differing intermolecular forces.
5. Illustrative Diagrams and Visuals
1. Graph of Raoult’s Law:
- A plot showing vapor pressure versus mole fraction for an ideal solution, with straight-line behavior.
2. Partial Vapor Pressure Contributions:
- Diagram showing how each component's vapor pressure varies with mole fraction, contributing to the total vapor pressure.
6. Sample Problems and Step-by-Step Solutions
Example Problem 1: Calculate the total vapor pressure of a solution with 0.4 mole fraction of component \( A \) (pure vapor pressure \( P_A^\circ = 100 \, \text{kPa}\)) and 0.6 mole fraction of component \( B \) (pure vapor pressure \( P_B^\circ = 80 \, \text{kPa}\)).
- Solution:
1. Calculate \( P_A \): \( P_A = P_A^\circ \cdot x_A = 100 \cdot 0.4 = 40 \, \text{kPa} \).
2. Calculate \( P_B \): \( P_B = P_B^\circ \cdot x_B = 80 \cdot 0.6 = 48 \, \text{kPa} \).
3. Total Vapor Pressure: \( P_{\text{total}} = P_A + P_B = 40 + 48 = 88 \, \text{kPa} \).
7. Common Tricks, Shortcuts, and Solving Techniques
- Direct Proportionality: For ideal solutions, use the mole fraction to calculate partial pressures directly without adjustments.
- Additive Pressures: In ideal solutions, simply sum partial pressures to find the total pressure.
8. Patterns in JEE Questions
JEE questions on Raoult's Law may involve:
- Calculating total vapor pressure of a solution given mole fractions and pure component vapor pressures.
- Analyzing graphs of vapor pressure vs. mole fraction to determine ideal or non-ideal behavior.
9. Tips to Avoid Common Mistakes
- Using Incorrect Mole Fractions: Ensure the correct mole fraction is used for each component.
- Misapplying Raoult’s Law to Non-Ideal Solutions: Raoult's Law applies strictly to ideal solutions; non-ideal solutions show deviations.
10. Key Points to Remember for Revision
- Raoult’s Law Formula: \( P_A = P_A^\circ \cdot x_A \) and \( P_B = P_B^\circ \cdot x_B \).
- Total Vapor Pressure: Sum of the partial pressures based on mole fractions.
- Applicability to Ideal Solutions Only: Deviations occur in non-ideal solutions.
11. Real-World Applications and Cross-Chapter Links
- Industrial Distillation: Raoult’s Law helps in designing distillation processes by predicting vapor pressures.
- Pharmaceutical Solutions: Used in formulating solutions and predicting behavior of volatile solvents in pharmaceuticals.
- Cross-Concept Connections: Related to colligative properties, which rely on vapor pressure changes in solutions.
Molecular Mass Determination Using Raoult's Law
1. Definition and Core Explanation
Raoult’s Law can be used to determine the molecular mass of a non-volatile solute in a solution. When a non-volatile solute is added to a volatile solvent, the vapor pressure of the solvent decreases in proportion to the mole fraction of the solute. This lowering of vapor pressure is a colligative property, depending on the number of solute particles rather than their identity.
The formula used to calculate molecular mass involves the relative lowering of vapor pressure, given by:
\[
\frac{P^\circ - P}{P^\circ} = \frac{\text{n}_{\text{solute}}}{\text{n}_{\text{solvent}}}
\]
where:
- \( P^\circ \) is the vapor pressure of the pure solvent,
- \( P \) is the vapor pressure of the solution,
- \( \text{n}_{\text{solute}} \) and \( \text{n}_{\text{solvent}} \) are the moles of solute and solvent, respectively.
Rearranging and substituting for moles using mass and molar mass, we can determine the molecular mass (\( M_{\text{solute}} \)) of the solute.
2. Steps for Molecular Mass Determination
- Measure Vapor Pressures: Obtain the vapor pressure of the pure solvent (\( P^\circ \)) and the vapor pressure of the solution (\( P \)).
- Calculate Relative Lowering of Vapor Pressure:
\[
\frac{P^\circ - P}{P^\circ}
\]
- Apply Mole Fraction Formula: Use the formula to express the mole fraction in terms of mass and molar mass:
\[
\frac{P^\circ - P}{P^\circ} = \frac{\text{mass of solute} / M_{\text{solute}}}{\text{mass of solvent} / M_{\text{solvent}}}
\]
- Solve for Molecular Mass: Rearrange to solve for \( M_{\text{solute}} \).
3. Key Terms and Concepts
- Relative Lowering of Vapor Pressure: The decrease in vapor pressure of the solvent due to the presence of a solute.
- Colligative Property: A property that depends on the concentration of solute particles rather than their identity, such as vapor pressure lowering.
- Non-Volatile Solute: A solute that does not vaporize and thus only affects the solvent’s vapor pressure.
4. Important Rules, Theorems, and Principles
- Raoult’s Law for Non-Volatile Solutes: States that the vapor pressure lowering depends on the mole fraction of the solute.
- Colligative Dependence: The molecular mass determination relies on the fact that vapor pressure lowering is a colligative property.
5. Illustrative Diagrams and Visuals
1. Vapor Pressure Lowering:
- Diagram showing the vapor pressure of the pure solvent compared to the vapor pressure of the solution after adding a non-volatile solute.
2. Molecular Mass Determination Setup:
- Schematic of an experiment where vapor pressures of pure solvent and solution are measured to calculate molecular mass.
[Include diagrams showing the decrease in vapor pressure when solute is added and the formula application for molecular mass calculation.]
6. Sample Problems and Step-by-Step Solutions
Example Problem 1: A solution is prepared by dissolving 2.5 g of a non-volatile solute in 80 g of water. The vapor pressure of pure water is 31.8 mm Hg, and the vapor pressure of the solution is 31.2 mm Hg. Calculate the molecular mass of the solute.
- Solution:
1. Calculate Relative Lowering of Vapor Pressure:
\[
\frac{P^\circ - P}{P^\circ} = \frac{31.8 - 31.2}{31.8} = 0.0189
\]
2. Convert Mass of Solvent to Moles:
\[
\text{Moles of water (solvent)} = \frac{80}{18} = 4.44 \, \text{mol}
\]
3. Apply Formula for Molecular Mass:
\[
0.0189 = \frac{2.5 / M_{\text{solute}}}{4.44}
\]
Rearranging,
\[
M_{\text{solute}} = \frac{2.5}{0.0189 \times 4.44} = 30 \, \text{g/mol}
\]
Example Problem 2: Determine the molecular mass of a solute if 1.5 g of it is dissolved in 100 g of benzene, lowering the vapor pressure from 100 mm Hg to 98 mm Hg.
- Solution:
1. Calculate the relative lowering:
\[
\frac{100 - 98}{100} = 0.02
\]
2. Convert mass of benzene to moles:
\[
\text{Moles of benzene} = \frac{100}{78} = 1.28 \, \text{mol}
\]
3. Apply molecular mass formula:
\[
M_{\text{solute}} = \frac{1.5}{0.02 \times 1.28} = 58.6 \, \text{g/mol}
\]
7. Common Tricks, Shortcuts, and Solving Techniques
- Directly Substitute Masses: Use given masses of solute and solvent along with their molar masses to simplify calculations.
- Remember Units for Consistency: Ensure all masses and pressures are in consistent units before calculating.
8. Patterns in JEE Questions
JEE Advanced questions often test:
- Calculating molecular mass using the relative lowering of vapor pressure.
- Recognizing relationships between molecular mass, vapor pressure lowering, and colligative properties.
9. Tips to Avoid Common Mistakes
- Misapplying Raoult’s Law: Ensure Raoult’s Law is only applied for non-volatile solutes in volatile solvents.
- Incorrect Mole Fractions: When calculating, use mole fractions appropriately and double-check mass-to-mole conversions.
10. Key Points to Remember for Revision
- Relative Lowering of Vapor Pressure Formula: \(\frac{P^\circ - P}{P^\circ} = \frac{\text{n}_{\text{solute}}}{\text{n}_{\text{solvent}}}\).
- Colligative Property Basis: Molecular mass determination via Raoult’s Law relies on the colligative nature of vapor pressure lowering.
- Importance of Non-Volatile Solutes: Only non-volatile solutes are used in molecular mass determination to avoid added vapor pressure.
11. Real-World Applications and Cross-Chapter Links
- Pharmaceuticals: Molecular mass determination is crucial in drug development for identifying active compounds.
- Industrial Solutions: Used in formulating precise mixtures where specific solute properties are required.
- Cross-Concept Connections: Related to other colligative properties like boiling point elevation and freezing point depression, which also depend on molecular mass.
Positive and Negative Deviations from Raoult's Law
1. Definition and Core Explanation
In real solutions, interactions between molecules often differ from those in ideal solutions, leading to deviations from Raoult’s Law. These deviations occur when intermolecular forces between solute and solvent molecules are either stronger or weaker than those in the pure components.
- Positive Deviations: Occur when the solute-solvent interactions are weaker than the solvent-solvent or solute-solute interactions. This leads to an increase in vapor pressure above the expected value from Raoult’s Law.
- Negative Deviations: Occur when the solute-solvent interactions are stronger than those in the pure components. This results in a decrease in vapor pressure below the value predicted by Raoult’s Law.
Examples:
- Positive Deviation: A mixture of ethanol and acetone exhibits a positive deviation because the hydrogen bonding in pure ethanol is disrupted by the addition of acetone, weakening the overall interactions.
- Negative Deviation: A mixture of water and nitric acid shows a negative deviation because strong hydrogen bonds form between water and nitric acid molecules, making it more difficult for them to vaporize.
2. Characteristics of Positive and Negative Deviations
- Positive Deviation:
- Higher vapor pressure than predicted by Raoult’s Law.
- Solute-solvent interactions are weaker than interactions in pure components.
- Enthalpy of mixing (\(\Delta H_{\text{mix}}\)) is positive (endothermic).
- Volume of mixing (\(\Delta V_{\text{mix}}\)) is positive (expansion occurs).
- Example: Ethanol and acetone, benzene and methanol.
- Negative Deviation:
- Lower vapor pressure than predicted by Raoult’s Law.
- Solute-solvent interactions are stronger than in the pure components.
- Enthalpy of mixing (\(\Delta H_{\text{mix}}\)) is negative (exothermic).
- Volume of mixing (\(\Delta V_{\text{mix}}\)) is negative (contraction occurs).
- Example: Water and hydrochloric acid, acetone and chloroform.
3. Key Terms and Concepts
- Deviation from Ideal Behavior: Occurs when actual vapor pressure differs from that predicted by Raoult’s Law.
- Endothermic Mixing: In positive deviations, energy is absorbed to overcome weaker solute-solvent interactions.
- Exothermic Mixing: In negative deviations, energy is released as stronger solute-solvent interactions form.
- Enthalpy of Mixing (\(\Delta H_{\text{mix}}\)): Heat absorbed or released when two substances are mixed, indicating the strength of interactions.
4. Important Rules, Theorems, and Principles
- Raoult’s Law Validity for Ideal Solutions: Deviations only occur in non-ideal solutions where intermolecular forces differ.
- Strength of Interactions and Deviation Type:
- Weaker solute-solvent interactions lead to positive deviations.
- Stronger solute-solvent interactions cause negative deviations.
5. Illustrative Diagrams and Visuals
1. Vapor Pressure Curve for Ideal vs. Non-Ideal Solutions:
- Show a graph with vapor pressure versus mole fraction, with ideal behavior as a straight line and positive/negative deviations curving above or below the line.
2. Molecular Interaction Diagrams:
- Illustration showing weakened interactions in positive deviations and strengthened interactions in negative deviations.
6. Sample Problems and Step-by-Step Solutions
Example Problem 1: Explain why a mixture of ethanol and water exhibits a negative deviation from Raoult’s Law.
- Solution:
1. Identify Interactions: Water and ethanol form strong hydrogen bonds with each other.
2. Effect on Vapor Pressure: These strong interactions make it harder for molecules to escape as vapor, lowering the vapor pressure.
- Conclusion: The mixture exhibits a negative deviation due to strong hydrogen bonding between ethanol and water.
Example Problem 2: A solution of benzene and methanol shows a vapor pressure higher than predicted. Identify the type of deviation and explain.
- Solution:
1. Interactions: Benzene disrupts the hydrogen bonding in methanol, leading to weaker solute-solvent interactions.
2. Effect on Vapor Pressure: Weaker interactions make it easier for molecules to vaporize, increasing the vapor pressure.
- Conclusion: This is a positive deviation from Raoult’s Law.
7. Common Tricks, Shortcuts, and Solving Techniques
- Identify Interaction Strength: Look at hydrogen bonding, polarity, and molecular size to estimate if interactions are stronger or weaker in the mixture.
- Enthalpy of Mixing Sign: A positive \(\Delta H_{\text{mix}}\) (endothermic) usually indicates positive deviation, while a negative \(\Delta H_{\text{mix}}\) (exothermic) indicates negative deviation.
8. Patterns in JEE Questions
JEE questions on Raoult's Law deviations often involve:
- Predicting if a given mixture will show positive or negative deviation.
- Analyzing molecular interactions to determine the deviation type.
- Understanding the impact of deviations on vapor pressure and enthalpy changes.
9. Tips to Avoid Common Mistakes
- Ignoring Intermolecular Forces: Always consider solute-solvent interactions compared to pure components to predict deviations.
- Misinterpreting Enthalpy of Mixing: Remember, a positive \(\Delta H_{\text{mix}}\) indicates weaker interactions (positive deviation), while a negative \(\Delta H_{\text{mix}}\) suggests stronger interactions (negative deviation).
10. Key Points to Remember for Revision
- Positive Deviations: Weaker solute-solvent interactions, higher vapor pressure, positive \(\Delta H_{\text{mix}}\).
- Negative Deviations: Stronger solute-solvent interactions, lower vapor pressure, negative \(\Delta H_{\text{mix}}\).
- Vapor Pressure Graph: Ideal solutions have a straight line; positive deviations curve above, and negative deviations curve below.
11. Real-World Applications and Cross-Chapter Links
- Azeotropes: Solutions with strong deviations from Raoult’s Law can form azeotropes, which have constant boiling points.
- Industrial Solvent Blending: Knowledge of deviations helps in predicting solubility, miscibility, and vapor pressure of mixtures in various industries.
- Cross-Concept Connections: Linked to colligative properties, which rely on vapor pressure changes, and thermodynamics due to enthalpy and entropy changes.
Characteristics of Ideal Solutions
1. Definition and Core Explanation
An ideal solution is a solution that obeys Raoult’s Law across all concentrations. In ideal solutions, the intermolecular forces between different molecules (solute-solvent) are similar to those between molecules of the same type (solute-solute and solvent-solvent). As a result, the physical properties of an ideal solution, such as vapor pressure, enthalpy of mixing, and volume of mixing, behave predictably and show no significant deviations.
Ideal solutions are characterized by having no enthalpy change or volume change upon mixing, as the energy and space requirements of the intermolecular forces remain constant.
2. Characteristics of Ideal Solutions
- Obeys Raoult's Law: The vapor pressure of each component in an ideal solution is directly proportional to its mole fraction in the solution.
\[
P_A = P_A^\circ \cdot x_A \quad \text{and} \quad P_B = P_B^\circ \cdot x_B
\]
where \( P_A \) and \( P_B \) are the partial vapor pressures, \( P_A^\circ \) and \( P_B^\circ \) are the vapor pressures of the pure components, and \( x_A \) and \( x_B \) are the mole fractions.
- No Enthalpy Change of Mixing (\(\Delta H_{\text{mix}} = 0\)): There is no heat absorbed or released when the components are mixed, as the intermolecular forces between different molecules are similar to those between like molecules.
- No Volume Change on Mixing (\(\Delta V_{\text{mix}} = 0\)): The total volume of the solution is the sum of the volumes of the pure components, as there is no expansion or contraction.
- Uniform Intermolecular Forces: The interactions between solute and solvent molecules are identical to those within the solute and solvent separately.
Examples:
- A mixture of benzene and toluene behaves almost ideally due to the similar structure and intermolecular forces in both compounds.
- A mixture of hexane and heptane is also close to ideal, as both are nonpolar alkanes with similar molecular sizes.
3. Key Terms and Concepts
- Raoult's Law: States that the partial vapor pressure of each component in a solution is proportional to its mole fraction.
- Enthalpy of Mixing (\(\Delta H_{\text{mix}}\)): The heat absorbed or released when two substances are mixed; for ideal solutions, this is zero.
- Volume of Mixing (\(\Delta V_{\text{mix}}\)): The change in volume upon mixing; ideal solutions have no volume change.
4. Important Rules, Theorems, and Principles
- Raoult's Law Validity: For a solution to be ideal, it must follow Raoult’s Law across all compositions.
- Uniform Intermolecular Forces Principle: Ideal solutions have intermolecular forces between solute-solvent that are comparable to those within solute-solute and solvent-solvent, leading to no significant enthalpy or volume changes.
5. Illustrative Diagrams and Visuals
1. Vapor Pressure of Ideal Solution:
- Graph showing the linear relationship between vapor pressure and mole fraction in an ideal solution, with vapor pressures of individual components contributing to the total vapor pressure.
2. Enthalpy and Volume of Mixing:
- Illustration showing that there is no enthalpy or volume change when two ideal components are mixed.
[Include diagrams showing linear vapor pressure behavior in ideal solutions and no changes in enthalpy and volume on mixing.]
6. Sample Problems and Step-by-Step Solutions
Example Problem 1: Calculate the total vapor pressure of an ideal solution containing 0.3 mole fraction of component \( A \) (pure vapor pressure \( P_A^\circ = 90 \, \text{kPa}\)) and 0.7 mole fraction of component \( B \) (pure vapor pressure \( P_B^\circ = 60 \, \text{kPa}\)).
- Solution:
1. Calculate \( P_A \): \( P_A = P_A^\circ \cdot x_A = 90 \cdot 0.3 = 27 \, \text{kPa} \).
2. Calculate \( P_B \): \( P_B = P_B^\circ \cdot x_B = 60 \cdot 0.7 = 42 \, \text{kPa} \).
3. Total Vapor Pressure: \( P_{\text{total}} = P_A + P_B = 27 + 42 = 69 \, \text{kPa} \).
Example Problem 2: Explain why a mixture of hexane and heptane behaves nearly ideally.
- Solution:
1. Similarity in Structure and Intermolecular Forces: Both hexane and heptane are nonpolar alkanes with similar molecular sizes and van der Waals forces.
2. Application of Raoult’s Law: Since the interactions between hexane-heptane molecules are nearly identical to those within hexane or heptane, the solution follows Raoult's Law closely.
- Conclusion: The mixture of hexane and heptane behaves ideally due to similar intermolecular forces and structural similarity.
7. Common Tricks, Shortcuts, and Solving Techniques
- Direct Use of Raoult's Law: In ideal solutions, partial pressures can be directly calculated by multiplying the mole fraction and pure component vapor pressures.
- Identifying Ideal Behavior: If components are structurally similar and have similar intermolecular forces, they likely form an ideal solution.
8. Patterns in JEE Questions
JEE Advanced questions on ideal solutions may involve:
- Calculating the total vapor pressure of ideal solutions using Raoult’s Law.
- Understanding the enthalpy and volume changes (or lack thereof) in ideal solutions.
- Identifying ideal or near-ideal behavior in given mixtures.
9. Tips to Avoid Common Mistakes
- Misapplying Raoult's Law: Ensure Raoult’s Law is applied only for ideal solutions or when the question specifies ideal behavior.
- Ignoring Mole Fraction Changes: When calculating partial pressures, ensure mole fractions are accurate and correspond to the correct components.
10. Key Points to Remember for Revision
- Raoult's Law for Ideal Solutions: \( P_A = P_A^\circ \cdot x_A \) and \( P_B = P_B^\circ \cdot x_B \).
- No Enthalpy and Volume Changes: Ideal solutions have \(\Delta H_{\text{mix}} = 0\) and \(\Delta V_{\text{mix}} = 0\).
- Characteristics of Ideal Behavior: Occurs when solute-solvent interactions are similar to those within pure components.
11. Real-World Applications and Cross-Chapter Links
- Distillation of Hydrocarbon Mixtures: Ideal solution behavior aids in the separation of hydrocarbons like hexane and heptane.
- Formulation of Solvents in Industry: Ideal solutions help predict solvent behavior, especially in nonpolar mixtures.
- Cross-Concept Connections: Related to colligative properties and deviations from ideality in non-ideal solutions.
|
Q 1. An azeotropic mixture of two liquids has boiling point lower than either of them when it: ; |
|
(A) Shows a negative deviation from Raoult's law; |
|
(B) Shows no deviation form Raoult's law; |
|
(C) Shows positive deviation from Raoult's law; |
|
(D) Is saturated; |
|
Q 2. Value of Henry's constant \(\mathrm{K}_{\mathrm{H}}\); |
|
(A) increases with increase in temperature; |
|
(B) decreases with increase in temperature; |
|
(C) remains constant.; |
|
(D) first increases then decreases.; |
|
Q 3. There are following the three solvents X, Y\&Z. They have identical molar masses. Match their boiling point with their \(\mathrm{k}_{\mathrm{b}}\)values:\n\begin{array}{ccc}\nSolvent & B. Pt. & \(\mathrm{K}_{\mathrm{b}}\) \\\n\(\mathrm{X}\) & \(100^{\circ} \mathrm{C}\) & 0.92 \\\n\(\mathrm{Y}\) & \(27^{\circ} \mathrm{C}\) & 0.63 \\\n\(\mathrm{Z}\) & \(283^{\circ} \mathrm{C}\) & 0.53 \\\n\end{array}\nAmong the following, which is the correct relation between the boiling point and \(\mathrm{K}_{\mathrm{b}}\) values for \(\mathrm{X}, \mathrm{Y}, \mathrm{Z}\) respectively?; |
|
(A) \(100^{\circ} \mathrm{C}-0.92,27^{\circ} \mathrm{C}-0.63,283^{\circ} \mathrm{C}-0.53\); |
|
(B) \(100^{\circ} \mathrm{C}-0.53,27^{\circ} \mathrm{C}-0.63,283^{\circ} \mathrm{C}-0.92\); |
|
(C) \(100^{\circ} \mathrm{C}-0.63,27^{\circ} \mathrm{C}-0.53,283^{\circ} \mathrm{C}-0.92\); |
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(D) \(100^{\circ} \mathrm{C}-0.53,27^{\circ} \mathrm{C}-0.92,283^{\circ} \mathrm{C}-0.63\); |
|
Q 4. Which one is a colligative property?; |
|
(A) Boiling point; |
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(B) Vapour pressure; |
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(C) Osmotic pressure; |
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(D) Freezing point; |
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Q 5. 7.6gmKBr in \(1250 \mathrm{ml}\) solution was found to have an osmotic pressure of \(1.804 \mathrm{~atm}\) at \(27^{\circ} \mathrm{C}\). Calculate degree of dissociation in \% (Atomic weight \(\mathrm{K}=39, \mathrm{Br}=80\) ); |
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(D) \(80 \%\); |
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(C) \(4.64 \%\); |
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(B) \(58.4 \%\); |
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(A) \(43.4 \%\); |
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Q 6. If molality of the dilute solution is doubled, the value of molal depression constant \(\left(\mathrm{K}_{\mathrm{f}}\right)\) will be; |
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(D) Unchanged; |
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(C) Tripled; |
|
(B) Halved; |
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(A) Doubled; |
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Q 7. Which of the following solutions contains the maximum freezing point?; |
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(D) \(0.001 \mathrm{M}\) Urea; |
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(C) \(0.01 \mathrm{M}\) Glucose; |
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(B) \(0.01 \mathrm{MBaCl}_{2}\); |
|
(A) \(0.01 \mathrm{MNaCl}\); |
|
Q 8. If all symbols have their usual meaning, then for a non-volatile non-electrolyte solute \(\mathrm{Lt}_{\mathrm{m} \rightarrow 1}\left(\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{m}}\right)\) is equal to:; |
|
(D) \(\mathrm{K}_{f}\); |
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(C) \(\Delta \mathrm{T}_{f}\); |
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(B) Infinity; |
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(A) Zero; |
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Q 9. The osmotic pressure of one molar solution at \(27^{\circ} \mathrm{C}\) is; |
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(D) \(12.1 \mathrm{~atm}\); |
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(C) \(1.21 \mathrm{~atm}\); |
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(B) \(24.6 \mathrm{~atm}\); |
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(A) \(2.46 \mathrm{~atm}\); |
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Q 10. A sample of drinking water was found to be severely contaminated with chloroform \(\left(\mathrm{CHCl}_{3}\right)\) supposed to a carcinogen. The level of contamination was \(15 \mathrm{ppm}\) (by mass): (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample.; |
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(D) \(6.45 \times 10^{-8} \mathrm{~m} 6.45 \times 10^{-8} \mathrm{~m}\); |
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(C) \(4.85 \times 10^{-6} \mathrm{~m}\); |
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(B) \(2.41 \times 10^{-4} \mathrm{~m}\); |
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(A) \(1.25 \times 10^{-4} \mathrm{~m}\); |
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Q 1. On the basis of information given below mark the correct option. Information:\n(i) In bromoethane and chloroethane mixture intermolecular interactions of \(\mathrm{A}-\mathrm{A}\) and \(\mathrm{B}-\mathrm{B}\) type are nearly same as A - B type interactions.\n(ii) In ethanol and acetone mixture, \(\mathrm{A}-\mathrm{A}\) or \(\mathrm{B}-\mathrm{B}\) type intermolecular interactions are stronger than \(\mathrm{A}-\mathrm{B}\) type interactions.\n(iii) In chloroform and acetone mixture, \(\mathrm{A}-\mathrm{A}\) or \(\mathrm{B}-\mathrm{B}\) type intermolecular interactions are weaker than \(\mathrm{A}-\mathrm{B}\) type interactions.; |
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(A) Solution (B) and (C) will follow Raoult's law.; |
|
(B) Solution (A) will follow Raoult's law.; |
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(C) Solution (B) will show negative deviation from Raoult's law.; |
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(D) Solution (C) will show positive deviation from Raoult's law.; |
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Q 1. For determination of molecular mass, Raoult's law is applicable only to; |
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(A) Dilute solutions of electrolytes; |
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(B) Concentrated solutions of electrolytes; |
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(C) Dilute solutions of non-electrolytes; |
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(D) Concentrated solution of non-electrolytes; |
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Q 1. When attraction between \(A-B\) is more than that of \(A-A\) and \(B-B\), the solution will show; |
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(A) Positive deviation from Raoult's law; |
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(B) Negative deviation from Raoult's law; |
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(C) No deviation from Raoult's law; |
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(D) Cannot be predicted; |
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Q 1. Which of the following is satisfied by an ideal solution?; |
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(A) Formation of an azeotropic mixture; |
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(B) \(\Delta \mathrm{S}_{\text {mix }}=0\); |
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(C) Raoult's law is obeyed under particular set of conditions only; |
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(D) \(\Delta \mathrm{H}_{\text {mix }}=0\); |
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Q 1. Heptane and octane form an ideal solution. At \(373 \mathrm{~K}\), the vapour pressures of the two liquid components are \(105.2 \mathrm{kPa}\) and \(46.8 \mathrm{kPa}\), respectively. What will be the vapour pressure of a mixture of \(26.0 \mathrm{~g}\) of heptane and \(35 \mathrm{~g}\) of octane?; |
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(A) \(7.308 \mathrm{kPa}\); |
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(B) \(73.08 \mathrm{kPa}\); |
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(C) \(730.8 \mathrm{kPa}\); |
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(D) \(7308 \mathrm{kPa}\); |
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Q 1. A sample of drinking water was found to be severely contaminated with chloroform \(\left(\mathrm{CHCl}_{3}\right)\) supposed to a carcinogen. The level of contamination was \(15 \mathrm{ppm}\) (by mass): (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample.; |
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(A) \(1.25 \times 10^{-4} \mathrm{~m}\); |
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(B) \(2.41 \times 10^{-4} \mathrm{~m}\); |
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(C) \(4.85 \times 10^{-6} \mathrm{~m}\); |
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(D) \(6.45 \times 10^{-8} \mathrm{~m} 6.45 \times 10^{-8} \mathrm{~m}\); |
Raoult's law for ideal solutions |
Molecular mass determination using Raoult's law |
Positive and negative deviations from Raoult's law |
Characteristics of ideal solutions |
Calculating partial vapor pressure in ideal solutions |
Colligative properties of solutions |
General concepts of solutions, concentration, and colligative properties |
Raoult's law - ideal solutions and deviations |
Molecular mass determination using Raoult's law for solutions |
Positive and negative deviations from Raoult's law in solutions |
Characteristics of ideal vs non-ideal solutions |
Partial vapor pressure calculations in binary mixtures |
Colligative properties - relative lowering of vapor pressure |