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#mole fraction

61 public questions tagged with this topic.

CHEMISTRY

What is the mole fraction of glucose in a solution containing 18 g of glucose (C₆H₁₂O₆) and 72 g of water? (Molar masses

Given: What is the mole fraction of glucose in a solution containing 18 g of glucose (C₆H₁₂O₆) and 72 g of water? (Molar masses: glucose = 180 g/mol, H₂O = 18 g/mol) These values define the system as per NCERT data. Formula: Moles of glucose = 18 / 180 = 0.1 mol. This is standard NCERT relation. Substitution & Calculation: Moles of H₂O = 72 / 18 = 4 mol. Total moles = 0.1 + 4 = 4.1. Mole fraction = 0.1 / 4.1 ≈ 0.0244. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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

A solution of two volatile liquids has vapor pressures of 400 mm Hg and 600 mm Hg for pure components. If the vapor pres

Raoult’s law: P = P₁⁰ · x₁ + P₂⁰ · (1 - x₁) . P = 400 × 0.6 + 600 × 0.4 = 240 + 240 = 480 mm Hg . Actual = 520 mm Hg ≠ 480 mm Hg, so it does not obey Raoult’s law.

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

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

A solution of two volatile liquids has vapor pressures of 250 mm Hg and 350 mm Hg for pure components. If the total vapo

Liquid phase: 310 = 250 x₁ + 350 (1 - x₁) . 310 = 250 x₁ + 350 - 350 x₁ , 100 x₁ = 40 , x₁ = 0.4 , x₂ = 0.6 . Vapor phase: y₁ = (P₁⁰ · x₁/Ptotal) = (250 × 0.4/310) ≈ 0.3226 .

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

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

A solution of two volatile liquids has vapor pressures of 600 mm Hg and 800 mm Hg. If the total vapor pressure is 680 mm

Liquid phase: 680 = 600 x₁ + 800 (1 - x₁) . 680 = 600 x₁ + 800 - 800 x₁ , 200 x₁ = 120 , x₁ = 0.6 , x₂ = 0.4 . Vapor phase: y₁ = (600 × 0.6/680) ≈ 0.5294 .

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

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