CUET UG Categorised PYQ Chemistry Unit 1

(Detailed) → According to Henry's Law: → p = KH × x → At constant pressure, lower KH gives higher mole fraction of gas in solution. → Higher mole fraction means greater solubility in water. → Methanal has KH = 0.000018, which is the lowest value among all options. → Hence, Methanal has maximum solubility in water.

(Detailed) → The ebullioscopic constant tells how much the boiling point of a solvent increases when a non-volatile solute is dissolved in it. → It depends on the nature of the solvent, molar mass, boiling point and enthalpy of vaporisation. → The relation is: → Kb = (R × M × Tb²) / (1000 × ΔHvap) → From standard comparative values of common solvents, carbon disulphide has the lowest ebullioscopic constant among the given options. → Hence, Carbon disulphide is the correct answer.

(Detailed) → Molal elevation constant is the elevation in boiling point produced when one mole of a non-volatile solute is dissolved in 1 kg of solvent. → Since it is connected with boiling point elevation, it is known as the ebullioscopic constant. → The word "ebullioscopic" is related to boiling. → Hence, Ebullioscopic constant is the correct answer.

(Detailed) → Osmotic pressure is given by: → π = CRT → At the same temperature, osmotic pressure depends directly on concentration. → Concentration of (A) = 0.5 mol / 1 L = 0.5 M → Concentration of (B) = 0.25 mol / 1 L = 0.25 M → Concentration of (C) = 0.1 mol / 0.01 L = 10 M → Concentration of (D) = 0.2 mol / 0.05 L = 4 M → Increasing order of concentration is: → 0.25 M < 0.5 M < 4 M < 10 M → Therefore, increasing order of osmotic pressure is: → (B) < (A) < (D) < (C) → Hence, option C is correct.

(Detailed) → For solutions of equal molarity, elevation in boiling point depends on the number of solute particles produced. → C₆H₁₂O₆ does not dissociate, so i = 1. → NaCl dissociates into Na⁺ and Cl⁻, so i = 2. → MgCl₂ dissociates into Mg²⁺ and 2Cl⁻, so i = 3. → AlCl₃ dissociates into Al³⁺ and 3Cl⁻, so i = 4. → Al₂(SO₄)₃ dissociates into 2Al³⁺ and 3SO₄²⁻, so i = 5. → Therefore, increasing order of elevation in boiling point is: → (A) < (B) < (C) < (D) < (E) → Hence, option A is correct.

(Detailed) → Molar mass of NaOH = 40 g mol⁻¹ → Given mass of NaOH = 5 g → Moles of NaOH = 5 / 40 = 0.125 mol → Volume of solution = 450 mL = 0.45 L → Molarity = moles / volume in litres → Molarity = 0.125 / 0.45 → Molarity = 0.2777 M ≈ 0.278 M → Hence, 0.278 M is the correct answer.

(Detailed) → Henry's law states that the partial pressure of a gas above a solution is directly proportional to its mole fraction in the solution. → Mathematically, it is written as: → P = KHx → Here, P is the partial pressure of the gas. → x is the mole fraction of the gas in solution. → KH is Henry's law constant. → Hence, Henry's law constant is the correct answer.

(Detailed) → Formula used: → ΔTf = i × Kf × m → Molar mass of benzoic acid = 122 g mol⁻¹ → Mass of benzoic acid = 2 g → Mass of benzene = 25 g → Molality, m = (Mass of solute / Molar mass) × (1000 / Mass of solvent in g) → m = (2 / 122) × (1000 / 25) → m = 0.6557 mol kg⁻¹ → Now, ΔTf = i × Kf × m → 1.62 = i × 4.9 × 0.6557 → i = 1.62 / (4.9 × 0.6557) → i ≈ 0.504 → Hence, 0.504 is the correct answer.

(Detailed) → In the Ostwald–Walker method, the relative lowering of vapour pressure is related to the mole fraction of solute. → The loss in mass of the water bulb and gain in mass of the CaCl₂ tube help determine vapour pressure lowering. → Using the given values in the relative lowering of vapour pressure relation gives the molecular mass of the organic compound. → According to the accepted calculation and answer key, the molecular mass is 200 g mol⁻¹. → Hence, 200 g mol⁻¹ is the correct answer.

(Detailed) → Acetic acid dissociates as: → CH₃COOH ⇌ CH₃COO⁻ + H⁺ → For dissociation into two ions: → i = 1 + α → Since Ka = 1.8 × 10⁻⁶ is very small, the degree of dissociation α is also very small. → Therefore, the van't Hoff factor of acetic acid will be close to 1. → It cannot be equal to 2 because complete dissociation is not taking place. → Hence, statement C is not correct.

(Detailed) → Statement (B) is correct because endothermic dissolution absorbs heat. → According to Le Chatelier's principle, if temperature is increased, the equilibrium shifts in the direction that absorbs heat. → Therefore, for endothermic dissolution, increase in temperature increases solubility. → Statement (A) is incorrect because gas solubility increases with increase in pressure, not decrease. → Statement (C) is incorrect because endothermic dissolution shows increase in solubility with temperature. → Statement (D) is incorrect because solubility of exothermic dissolution is affected by temperature. → Statement (E) is incorrect because exothermic dissolution generally decreases solubility with increase in temperature. → Hence, (B) only is the correct answer.

(Detailed) → In an ideal solution, the intermolecular interactions between unlike molecules are nearly the same as those between like molecules. → Therefore, there is no heat change during mixing. → So, ΔmixH = 0. → Also, there is no volume change during mixing. → So, ΔmixV = 0. → Since no heat is exchanged with the surroundings, ΔSsurrounding = 0. → Mixing increases randomness of the system, so ΔSsystem is positive. → Statement (A) is incorrect because spontaneous mixing has negative ΔG, not positive ΔG. → Hence, (B), (C), (D), (E) only is the correct answer.

(Detailed) → Mass of glucose dissolved = 4.5 g → Molar mass of glucose = 180 g mol⁻¹ → Moles of glucose = 4.5 / 180 → Moles of glucose = 0.025 mol → Volume of solution = 40 mL = 0.040 L → Molarity = Moles of solute / Volume of solution in litres → Molarity = 0.025 / 0.040 → Molarity = 0.625 M → Hence, 0.625 M is the correct answer.

(Detailed) → Molality is defined as the number of moles of solute dissolved per kilogram of solvent. → It is based on mass, not volume. → Mass remains unaffected by temperature changes. → Therefore, molality does not change when temperature changes. → Molarity, strength and % (w/v) depend on volume, and volume changes with temperature. → Hence, molality is the most reliable concentration term.

(Detailed) → For dissociation, van't Hoff factor is calculated by: → i = 1 + α(n − 1) → For K₂SO₄: → n = 3, α = 0.6 → i = 1 + 0.6(3 − 1) → i = 1 + 1.2 = 2.2 → So, (A)-(III). → For K₃Fe(CN)₆: → n = 4, α = 0.9 → i = 1 + 0.9(4 − 1) → i = 1 + 2.7 = 3.7 → So, (B)-(I). → For AlCl₃: → n = 4, α = 0.8 → i = 1 + 0.8(4 − 1) → i = 1 + 2.4 = 3.4 → So, (C)-(IV). → For K₂HgI₄: → n = 3, α = 0.4 → i = 1 + 0.4(3 − 1) → i = 1 + 0.8 = 1.8 → So, (D)-(II). → Therefore, correct matching is: → (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

(Detailed) → Given molality = 0.25 m → Mass of solvent = 2.5 kg → Moles of urea = Molality × Mass of solvent in kg → Moles of urea = 0.25 × 2.5 → Moles of urea = 0.625 mol → Molar mass of urea, NH₂CONH₂ = 60 g mol⁻¹ → Mass of urea = Moles × Molar mass → Mass of urea = 0.625 × 60 → Mass of urea = 37.5 g → Hence, 37.5 g is the correct answer.

(Detailed) → Solutions can be classified based on the physical states of solute and solvent. → Camphor exists in solid state. → Nitrogen acts as the gaseous medium. → Therefore, camphor in nitrogen gas represents a solid dispersed in gas type system. → Hence, Solid – Gas is the correct answer.

(Detailed) → Statement (A) is correct because osmotic pressure method is suitable for determining molar mass of proteins and other macromolecules. → Statement (B) is correct because osmotic pressure follows the formula: → π = CRT → Therefore, osmotic pressure is directly proportional to molarity. → Statement (C) is correct because reverse osmosis occurs when pressure greater than osmotic pressure is applied on the concentrated solution side. → Statement (D) is correct because edema involves retention of water in tissue cells due to osmotic imbalance. → Therefore, all four statements are correct. → Hence, (A), (B), (C) and (D) is the correct answer.

(Detailed) → According to Raoult's law: → Ptotal = xA PA° + xD PD° → Given: → PA° = 500 mm Hg → PD° = 800 mm Hg → Ptotal = 700 mm Hg → Since xA + xD = 1, xA = 1 − xD. → Substitute in the equation: → 700 = (1 − xD)(500) + xD(800) → 700 = 500 − 500xD + 800xD → 700 = 500 + 300xD → 300xD = 200 → xD = 200 / 300 = 2/3 = 0.6667 → Mole percentage of D = 0.6667 × 100 → Mole percentage of D = 66.67% → Hence, 66.67 mole percent is the correct answer.

(Detailed) → For association of solute molecules, the formula is: → i = 1 − α + α/n → Given: → i = 0.80 → α = 0.3 → Substitute the values: → 0.80 = 1 − 0.3 + 0.3/n → 0.80 = 0.70 + 0.3/n → 0.10 = 0.3/n → n = 0.3 / 0.10 → n = 3 → Therefore, the value of n is 3. → Hence, 3 is the correct answer.

(Detailed) → A solid solution is a homogeneous mixture in which the solvent is in the solid state. → In many metallic solutions, one metal dissolves in another metal to form an alloy-like mixture. → Amalgams are solutions involving mercury and another metal. → Amalgam of mercury with sodium is treated as a solid solution in this context. → Therefore, amalgam of mercury with sodium is the correct example of solid solution.

(Detailed) → Henry's law states that the partial pressure of a gas above a solution is directly proportional to its mole fraction in the solution. → The mathematical expression is: → P = KHx → Here, P is the partial pressure of the gas. → x is the mole fraction of the gas in solution. → KH is Henry's Law constant. → Therefore, KH represents Henry's Law constant.

(Detailed) → Under high pressure in deep water, more nitrogen dissolves in the blood of divers. → When a diver comes up too quickly, the external pressure decreases suddenly. → Due to this sudden decrease in pressure, dissolved nitrogen comes out of the blood in the form of bubbles. → These bubbles may block capillaries and cause severe pain and medical complications. → This condition is called bends or decompression sickness. → Hence, Bends is the correct answer.

(Detailed) → Ideal solutions obey Raoult's law over the entire range of composition. → Nearly ideal solutions are formed when the components have similar size, structure and intermolecular forces. → n-hexane and n-heptane are similar non-polar hydrocarbons, so they behave nearly ideally. → Benzene and toluene are also similar aromatic liquids and behave nearly ideally. → Bromoethane and chloroethane are similar haloalkanes, so they are closer to ideal behaviour. → Ethanol and acetone have different types of intermolecular interactions. → Ethanol shows hydrogen bonding, while acetone mainly shows dipole-dipole interactions. → Due to this difference, ethanol and acetone do not fall under nearly ideal solution category. → Hence, Ethanol and acetone is the correct answer.