Skip to content
Login Register

#colligative properties

71 public questions tagged with this topic.

CHEMISTRY

A solution of 2.44 g of acetic acid ( CH₃COOH ) in 100 g of benzene shows a freezing point depression of 0.5 K. If the

Given: A solution of 2.44 g of acetic acid ( CH₃COOH ) in 100 g of benzene shows a freezing point depression of 0.5 K. If the molar mass is 60 g/mol, what is the van't Hoff factor due to dimerization? ( K_f = 5.12 K kg/mol ) These values define the system as per NCERT data. Formula: Molality = 2.44 / 60/0.1 = 0.4067 mol/kg. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Normal Δ T_f = 5.12 × 0.4067 = 2.08 K . Observed Δ T_f = 0.5 K . i = 0.5/2.08 = 0.24 . Result: The computed value matches the expected outcome and confirms the correct c

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.

View Details
CHEMISTRY

The boiling point elevation of a 0.15 m aqueous KCl solution is 0.156 K. What is the van't Hoff factor? ( K_b = 0.52 K k

Given: The boiling point elevation of a 0.15 m aqueous KCl solution is 0.156 K. What is the van't Hoff factor? ( K_b = 0.52 K kg/mol ) These values define the system as per NCERT data. Formula: Δ T_b = i · K_b · m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 0.156 = i × 0.52 × 0.15 . i = 0.156/0.52 × 0.15 = 0.156/0.078 = 2 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Solutions, Topic: Concentration terms and colligative properties.

View Details
CHEMISTRY

A solution of 1.8 g of benzoic acid ( C₆H₅COOH ) in 30 g of benzene shows a freezing point depression of 1.2 K. What is

Given: A solution of 1.8 g of benzoic acid ( C₆H₅COOH ) in 30 g of benzene shows a freezing point depression of 1.2 K. What is the van't Hoff factor if it dimerizes? ( K_f = 5.12 K kg/mol, Molar mass = 122 g/mol ) Formula: Molality = 1.8 / 122/0.03 = 0.492 mol/kg. Substitution & Calculation: Normal Δ T_f = 5.12 × 0.492 = 2.52 K . Observed Δ T_f = 1.2 K . i = 1.2/2.52 = 0.476 . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

View Details
CHEMISTRY

A solution of 6.8 g of a non-volatile solute in 200 g of water has a boiling point elevation of 0.26 K. What is the mola

Given: A solution of 6.8 g of a non-volatile solute in 200 g of water has a boiling point elevation of 0.26 K. What is the molar mass of the solute? ( K_b = 0.52 K kg/mol ) These values define the system as per NCERT data. Formula: Molality = Δ T_b/K_b = 0.26/0.52 = 0.5 mol/kg. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles = 0.5 × 0.2 = 0.1 mol . Molar mass = 6.8/0.1 = 68 g/mol . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are proper

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details
CHEMISTRY

A solution of 4.8 g of a non-volatile solute in 150 g of water has a boiling point elevation of 0.208 K. What is the mol

Given: A solution of 4.8 g of a non-volatile solute in 150 g of water has a boiling point elevation of 0.208 K. What is the molar mass of the solute? ( K_b = 0.52 K kg/mol ) These values define the system as per NCERT data. Formula: Molality = Δ T_b/K_b = 0.208/0.52 = 0.4 mol/kg. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles = 0.4 × 0.15 = 0.06 mol . Molar mass = 4.8/0.06 = 80 g/mol . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are p

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details
CHEMISTRY

How many grams of ethylene glycol ( C₂H₆O₂ ) are required to depress the freezing point of 300 g of water by 1.24

Given: How many grams of ethylene glycol ( C₂H₆O₂ ) are required to depress the freezing point of 300 g of water by 1.24 K? ( K_f = 1.86 K kg/mol, Molar mass of ethylene glycol = 62 g/mol ) These values define the system as per NCERT data. Formula: Δ T_f = K_f · m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 1.24 = 1.86 · m, m = 0.667 mol/kg . Moles = 0.667 × 0.3 = 0.2 mol . Mass = 0.2 × 62 = 12.4 g . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹,

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details
CHEMISTRY

A solution of 12 g of a non-volatile solute in 100 g of water boils at 100.52°C at 1 atm. What is the molar mass of the

Given: A solution of 12 g of a non-volatile solute in 100 g of water boils at 100.52°C at 1 atm. What is the molar mass of the solute? ( K_b = 0.52 K kg/mol ) These values define the system as per NCERT data. Formula: Δ T_b = 100.52 - 100 = 0.52 K. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Molality = 0.52/0.52 = 1 mol/kg . Moles = 1 × 0.1 = 0.1 mol . Molar mass = 12/0.1 = 120 g/mol . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are pro

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details
Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

The vapor pressure of pure water is 28 mm Hg at a certain temperature. A solution with a non-volatile solute has a vapor

(p⁰ - p/p⁰) = xsolute . (28 - 26.6/28) = (1.4/28) = 0.05 . Moles of water = (360/18) = 20 . xsolute = (nsolute/nsolute + 20) = 0.05 . nsolute = 0.05 (nsolute + 20) , 0.95 nsolute = 1 , nsolute ≈ 1.0526 . Mass = 1.0526 × 50 ≈ 52.63 g .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

View Details
Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

A solution of a non-volatile solute in water has a vapor pressure of 22.4 mm Hg at a temperature where pure water’s vapo

Δ Tb = 100.208 - 100 = 0.208 K . Δ Tb = Kb · m . 0.208 = 0.52 · m , m = (0.208/0.52) = 0.4 mol/kg . Cross-check: xsolute = (24 - 22.4/24) = 0.0667 , consistent for dilute solution.

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

View Details