13 public questions tagged with this topic.
kappa = (cell constant/R) = (0.4/250) = 0.0016 S cm⁻¹ . Lambdam = (kappa × 1000/c) = (0.0016 × 1000/0.02) = 80 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
kappa = (cell constant/R) = (0.4/250) = 0.0016 S cm⁻¹ . Lambdam = (kappa × 1000/c) = (0.0016 × 1000/0.02) = 80 S cm² mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Conductance - Conductivity Molar Conductivity and Kohlrausch Law
Moles of NaCl = 0.25 × 0.4 = 0.1 mol . Mass = 0.1 × 58.5 = 5.85 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions
Δ Tb = i · Kb · m . Δ Tb = 1.87 × 0.52 × 0.1 ≈ 0.097 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Relative Lowering and Elevation of Boiling Point
Pi = i · M · RT . Molarity ≈ molality for dilute solution, so M = 0.05 M . Pi = 1.9 × 0.05 × 0.0821 × 300 ≈ 2.339 atm .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Δ Tf = i · Kf · m . Δ Tf = 1.9 × 1.86 × 0.25 ≈ 0.8835 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Depression of Freezing Point
Δ Tf = i · Kf · m . 1.116 = 2 × 1.86 × (moles/0.5) . 1.116 = 3.72 × (moles/0.5) , moles = (1.116 × 0.5/3.72) ≈ 0.15 . Mass = 0.15 × 58.5 = 8.775 g .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Abnormal Molar Masses and van't Hoff Factor
Moles of NaCl = 4.8/58.5 ≈ 0.0821 mol. Volume = 0.25 L. Molarity = 0.0821/0.25 ≈ 0.328 M.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Reactions in Solutions and Mass Percentage and Numericals
M₁V₁ = M₂V₂. 0.2 × 150 = M₂ × (150 + 250). M₂ = (0.2 × 150)/400 = 0.075 M.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Laws of Chemical Combination - Conservation Mass Definite Multiple
Mass of 1 L solution = 1000 × 1.25 = 1250 g. Mass of NaCl = 3 × 58.5 = 175.5 g. Mass of water = 1250 − 175.5 = 1074.5 g = 1.0745 kg. Molality = 3 / 1.0745 ≈ 2.79 m.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Laws of Chemical Combination - Conservation Mass Definite Multiple
Total mass = 5.85 + 44.15 = 50 g. Mass % = (5.85/50) × 100 = 11.7%.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Importance and Nature of Chemistry and Properties of Matter
Moles of NaCl = 11.7/58.5 = 0.2 mol. Volume = 300 mL = 0.3 L. Molarity = 0.2/0.3 ≈ 0.667 M.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Concentration Terms - Molarity Molality and Mole Fraction
