22 public questions tagged with this topic.
Given: m = 0.25kg, Lv = 8.5×105J kg−1. Q = mLv = 0.25×8.5×105 = 212500J = 212.5kJ. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 212.5 kJ. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.
h = 2Scosθρga. S = 0.0227N/m, ρ = 806kg/m3, g = 9.8m/s2, a = 0.25×10−3m. h = 2×0.0227×1806×9.8×0.25×10−3 = 0.02298m = 2.3cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.3 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
h = 2Scosθρga. S = 0.0227N/m, ρ = 806kg/m3, g = 10m/s2, a = 0.35×10−3m. h = 2×0.0227×1806×10×0.35×10−3 = 0.0161m = 1.61cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.6 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Ethanol (CH₃CH₂OH) is commonly used as a fuel additive (e.g., in gasoline) to improve the octane rating and reduce emissions.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Preparation of Phenols and Ethers - Williamson and Other Methods
Ethanol is the most commonly used solvent in pharmaceutical preparations due to its effectiveness and relatively low toxicity.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Preparation of Phenols and Ethers - Williamson and Other Methods
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
Ethanol (CH₃CH₂OH) forms hydrogen bonds due to the -OH group, requiring more energy to break, whereas methoxymethane (CH₃OCH₃) lacks hydrogen bonding, leading to a lower boiling point.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Physical Properties and Acidity of Alcohols and Phenols
Glycerol (CH₂OH-CHOH-CH₂OH) has three -OH groups, forming extensive hydrogen bonding, which increases its boiling point compared to ethanol with one -OH group.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Classification and Nomenclature and Structures of Alcohols Phenols Ethers
Methanol (CH₃OH) has a lower molecular weight (32 g/mol) than ethanol (CH₃CH₂OH, 46 g/mol), reducing van der Waals forces and thus lowering its boiling point.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Chemical Reactions of Alcohols - Dehydration Oxidation Esterification
NaSH provides SH⁻ , which substitutes Br in CH₃CH₂Br via SN₂ , forming CH₃CH₂SH (ethanethiol).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Nature of C-X Bond and Methods of Preparation
CH₃CH₂CH₂Br reacts via SN₁ with AgNO₃ , forming AgBr (yellow precipitate) and CH₃CH₂CH₂ONO₂ .
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Classification and Nomenclature and Isomerism of Haloalkanes
Mole fraction of ethanol = 1 - 0.2 = 0.8 . Vapor pressure = 60 × 0.8 = 48 mm Hg .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration
