7 public questions tagged with this topic.
Aqueous KOH substitutes Cl in CH₃CHClCH₂CH₂CH₃ (2-chloropentane) via SN₂ to form 2-pentanol ( CH₃CH(OH)CH₂CH₂CH₃ ).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Nature of C-X Bond and Methods of Preparation
Reduction involves electron gain. O (0 in O₂) becomes -1 in H₂ (via intermediate steps), gaining electrons.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Types of Redox Reactions - Combination Decomposition Displacement
For KOH , [OH-] = 0.01 M = 10⁻² , pOH = -log(10⁻²) = 2 , pH = 14 - 2 = 12 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis
Reaction: KOH + HCl → KCl + H₂O. Moles of KCl = 7.45/74.5 = 0.1 mol. 1 mol KCl from 1 mol KOH; mass = 0.1 × 56 = 5.6 g.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
M₁V₁ = M₂V₂. 0.15 × 50 = M₂ × (50 + 150). M₂ = (0.15 × 50)/200 = 0.0375 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 × 1.1 = 1100 g. Mass of KOH = 2 × 56 = 112 g. Mass of water = 988 g = 0.988 kg. Molality = 2 / 0.988 ≈ 2.02 m.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Laws of Chemical Combination - Conservation Mass Definite Multiple
M₁V₁ = M₂V₂. 0.1 × 100 = M₂ × (100 + 200). M₂ = (0.1 × 100)/300 ≈ 0.0333 M.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Concentration Terms - Molarity Molality and Mole Fraction
