#solution chemistry

Conductance - Conductivity Molar Conductivity and Kohlrausch Law

The molar conductivity of a 0.005 M solution is 80 S cm² mol⁻¹. What is the conductivity of the solution?

Lambdam = (kappa × 1000/c) , 80 = (kappa × 1000/0.005) , kappa = 0.0004 S cm⁻¹ .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law

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

The vapor pressure of pure water at a certain temperature is 50 mm Hg. What is the vapor pressure of a solution with a s

Mole fraction of water = 1 - 0.1 = 0.9 . Vapor pressure = 50 × 0.9 = 45 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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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

The vapor pressure of pure methanol is 90 mm Hg. What is the vapor pressure of a solution with a solute mole fraction of

Mole fraction of methanol = 1 - 0.3 = 0.7 . Vapor pressure = 90 × 0.7 = 63 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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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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Types of Solutions and Expressing Concentration

A solution of 10 g of a solute (molar mass = 40 g/mol) in 400 g of water has a freezing point of -0.93°C. What is the va

Moles = (10/40) = 0.25 . Molality = (0.25/0.4) = 0.625 mol/kg . Δ Tf = i · Kf · m . 0.93 = i × 1.86 × 0.625 , i = (0.93/1.86 × 0.625) ≈ 0.8 .