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PUCET (Punjab) Solutions — practice questions — page 3

Explore 110 free Solutions questions in PUCET (Punjab) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Showing questions 61–90 of 110.

Freezing point depression: ΔT_f = K_f × m. For 0.5 molal NaCl (assume full dissociation, K_f = 1.86), the depression is:Read the answer and worked solution →Vapour pressure of pure water at 25°C is 23.8 mm Hg. The vapour pressure of a solution with 18 g glucose (M.W. = 180) in 90 g water:Read the answer and worked solution →Osmotic pressure π = nRT/V (or CRT for molarity C). What is π for 0.1 M sugar solution at 27°C (R = 0.0821)?Read the answer and worked solution →The van't Hoff factor i for a strong electrolyte like CaCl2 (full dissociation) is:Read the answer and worked solution →In ideal solutions, ΔH_mix is:Read the answer and worked solution →Mass percentage (w/w) of a solution: 12 g sugar in 100 g of solution is:Read the answer and worked solution →Two solutions are isotonic if:Read the answer and worked solution →Maximum boiling azeotropes are formed by mixtures with:Read the answer and worked solution →Molality of a solution does not depend on:Read the answer and worked solution →For 0.5 m solutions of NaCl, KCl, and BaCl2, which gives largest ΔT_f?Read the answer and worked solution →Osmotic pressure of 0.1 M Glucose at 27°C is π_g. For 0.1 M NaCl at same T, the osmotic pressure is approximately:Read the answer and worked solution →The pKa of a weak acid HA is 4.75. Find pH of an equimolar mixture of HA and A⁻:Read the answer and worked solution →Solubility of N2 in water at 25°C is 6.8 × 10⁻⁴ mol/L at 1 atm partial pressure. By Henry's law, at 5 atm partial pressure of N2:Read the answer and worked solution →Two solutions A (0.1 M urea, π_A) and B (0.1 M NaCl, π_B). Water flows by osmosis:Read the answer and worked solution →A solution boils at 100.5°C (water at 100). If K_b = 0.52, molality is:Read the answer and worked solution →Mole fraction of solute in 18% glucose (180 g/mol) by mass in water:Read the answer and worked solution →Vapour pressure of pure water is 100 (arbitrary units). Add 1 mole non-volatile solute to 9 moles water. New vapour pressure is:Read the answer and worked solution →A solution is:Read the answer and worked solution →Mole fraction of solute x_B in a binary solution:Read the answer and worked solution →Molarity M is defined as:Read the answer and worked solution →Molality m is defined as:Read the answer and worked solution →Henry's law states partial pressure of a gas above liquid is proportional to:Read the answer and worked solution →Raoult's law for ideal solution states vapor pressure of solvent:Read the answer and worked solution →For dilute solutions, lowering of vapor pressure of solvent ΔP / P°_A =Read the answer and worked solution →Elevation in boiling point ΔT_b for solution of non-volatile solute:Read the answer and worked solution →Depression in freezing point ΔT_f for non-electrolyte solute:Read the answer and worked solution →Osmotic pressure π for dilute solution:Read the answer and worked solution →For NaCl in water (assumes complete dissociation), van't Hoff factor i =Read the answer and worked solution →For 0.1 M urea in water, freezing point depression (K_f = 1.86):Read the answer and worked solution →Mass percent of solute = (mass of solute / mass of solution) × 100. For 5 g NaCl in 95 g water:Read the answer and worked solution →

Worked practice questions

Freezing point depression: ΔT_f = K_f × m. For 0.5 molal NaCl (assume full dissociation, K_f = 1.86), the depression is:

  1. 0.46 K
  2. 0.93 K
  3. 3.72 K
  4. 1.86 K
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Correct answer: D. 1.86 K

NaCl dissociates into 2 ions (i = 2 van't Hoff factor). ΔT_f = i × K_f × m = 2 × 1.86 × 0.5 = 1.86 K. So 0.5 m NaCl freezes at -1.86°C.

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Vapour pressure of pure water at 25°C is 23.8 mm Hg. The vapour pressure of a solution with 18 g glucose (M.W. = 180) in 90 g water:

  1. 23.1 mm Hg
  2. 23.3 mm Hg
  3. 25 mm Hg
  4. 22.4 mm Hg
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Correct answer: B. 23.3 mm Hg

Moles glucose = 18/180 = 0.1. Moles water = 90/18 = 5. Mole fraction water = 5/5.1 ≈ 0.98. P = x_water × P_pure = 0.98 × 23.8 ≈ 23.3 mm Hg. (Glucose lowers vapour pressure of water.)

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Osmotic pressure π = nRT/V (or CRT for molarity C). What is π for 0.1 M sugar solution at 27°C (R = 0.0821)?

  1. 2.46 atm
  2. 24.6 atm
  3. 0.246 atm
  4. 246 atm
Show answer and solution

Correct answer: A. 2.46 atm

π = CRT = 0.1 × 0.0821 × 300 ≈ 2.46 atm. Osmotic pressure is substantial even at modest concentrations.

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More Chemistry topics

Ideal and Non-Ideal SolutionsIdeal SolutionsNon-Ideal SolutionsReactions in Aqueous SolutionElectrochemistryElectrochemical SeriesCoordination CompoundsSurface Chemistry