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#boiling point elevation

23 public questions tagged with this topic.

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 properly used as per NCERT.

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.

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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

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