19 public questions tagged with this topic.
Propane-1,2-diol (CH₃CH(OH)CH₂OH) has two -OH groups, enhancing hydrogen bonding with water, making it more soluble than monohydric alcohols like propan-1-ol.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Preparation of Alcohols - from Alkenes Haloalkanes Carbonyls Grignard
Phenol is more acidic due to resonance stabilization of the phenoxide ion, allowing it to form a soluble salt with NaOH, unlike ethanol, which is a weaker acid.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Physical Properties and Acidity of Alcohols and Phenols
Crystallization uses differential solubility (e.g., ethanol) to dissolve one solid and recrystallize it, leaving the insoluble one behind, ideal when melting points are irrelevant.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Purification Methods - Crystallization Distillation Chromatography
Sublimation relies on sublimation temperature differences, ineffective here. Crystallization or extraction using water solubility would work instead.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Purification Methods - Crystallization Distillation Chromatography
For CaCO₃ Ca²⁺ + CO₃²⁻ , Ksp = [Ca²⁺][CO₃²⁻] = 3.8 × 10⁻⁹ , [CO₃²⁻] ≈ 0.02 , [Ca²⁺] = (3.8 × 10⁻⁹/0.02) = 1.9 × 10⁻⁷ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Ksp = [Hg₂²⁺][Cl-]² = 1.3 × 10⁻¹⁸ , [Cl-] ≈ 0.01 , [Hg₂²⁺] (0.01)² = 1.3 × 10⁻¹⁸ , [Hg₂²⁺] = (1.3 × 10⁻¹⁸/0.0001) = 1.3 × 10⁻¹⁴ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Ksp = [Ag+][Cl-] = S² = 1.8 × 10⁻¹⁰ , so S = sqrt1.8 × 10⁻¹⁰ = 1.34 × 10⁻⁵ mol/L .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
For AgBr Ag+ + Br- , Ksp = [Ag+][Br-] = S² = (7.1 × 10⁻⁷)² = 5.04 × 10⁻¹³ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
For Ag₂S 2Ag+ + S²⁻ , [Ag+] = 4.0 × 10⁻¹⁶ , [S²⁻] ≈ 0.01 , Ksp = (4.0 × 10⁻¹⁶)² × 0.01 = 1.6 × 10⁻³³ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
Ksp = [X²⁺][Y-]² = S (2S)² = 4S³ = 2.7 × 10⁻⁷ , S³ = (2.7 × 10⁻⁷/4) = 6.75 × 10⁻⁸ , S ≈ 4.07 × 10⁻³ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
For Ag₂CrO₄ 2Ag+ + CrO₄²⁻ , [Ag+] = 2.0 × 10⁻⁵ , [CrO₄²⁻] ≈ 0.02 , Ksp = (2.0 × 10⁻⁵)² × 0.02 = 8.0 × 10⁻¹² .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
Ksp = [Ba²⁺][F-]² = 1.7 × 10⁻⁶ . [F-] = 0.01 + 2[Ba²⁺] ≈ 0.01 M (since [Ba²⁺ is small). [Ba²⁺] (0.01)² = 1.7 × 10⁻⁶ , [Ba²⁺] = (1.7 × 10⁻⁶/0.0001) = 1.7 × 10⁻² M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
