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#solution chemistry

43 public questions tagged with this topic.

CHEMISTRY

What is the mole fraction of NaCl in a solution containing 5.85 g of NaCl and 90 g of water? (Molar masses: NaCl = 58.5

Given: What is the mole fraction of NaCl in a solution containing 5.85 g of NaCl and 90 g of water? (Molar masses: NaCl = 58.5 g/mol, H₂O = 18 g/mol) These values define the system as per NCERT data. Formula: Moles of NaCl = 5.85 / 58.5 = 0.1 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles of H₂O = 90 / 18 = 5 mol. Total moles = 0.1 + 5 = 5.1. Mole fraction = 0.1 / 5.1 ≈ 0.0196. 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 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 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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Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

The vapor pressure of a solvent decreases from 50 mm Hg to 47 mm Hg when a non-volatile solute is added. If an additiona

(p⁰ - p/p⁰) = xsolute . Initial: (50 - 47/50) = 0.06 . Doubling molality doubles xsolute (approximately for dilute solutions), so new xsolute = 0.12 . New p = p⁰ (1 - xsolute) = 50 (1 - 0.12) = 44 mm Hg .

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

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