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PUCET (Punjab) Surface Chemistry — practice questions — page 2

Explore 106 free Surface Chemistry questions in PUCET (Punjab) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Surface Chemistry questions and exercises with answers

Showing questions 31–60 of 106.

For water, K_b = 0.52 K kg mol⁻¹. What is the boiling point of a solution of 18 g glucose (M = 180) in 1 kg water?Read the answer and worked solution →Depression of freezing point (ΔT_f) is given by:Read the answer and worked solution →45 g of ethylene glycol (M = 62) is dissolved in 600 g of water. K_f(water) = 1.86 K kg mol⁻¹. The depression in freezing point is approximately:Read the answer and worked solution →The osmotic pressure (Π) of a dilute solution is given by:Read the answer and worked solution →Two solutions are termed *isotonic* if they:Read the answer and worked solution →Blood cells placed in a hypertonic solution (e.g., > 0.9% NaCl) will:Read the answer and worked solution →Why do wilted flowers revive when placed in fresh water?Read the answer and worked solution →The van't Hoff factor (i) for a solute that completely dissociates into n particles is:Read the answer and worked solution →For benzoic acid dimerising in benzene (2 CH₃COOH ⇌ (CH₃COOH)₂), the van't Hoff factor approaches:Read the answer and worked solution →Inclusion of the van't Hoff factor modifies the boiling point elevation formula to:Read the answer and worked solution →Experimentally, a solute is found to have a molar mass *lower* than its formula mass. This indicates the solute is:Read the answer and worked solution →In an antifreeze formulation, 35% (v/v) ethylene glycol in water lowers the freezing point of water to about:Read the answer and worked solution →Adsorption isotherm:Read the answer and worked solution →Hydrogenation of vegetable oils to make margarine uses catalyst:Read the answer and worked solution →According to VSEPR theory, the shape of a molecule is determined byRead the answer and worked solution →The hybridisation of carbon in methane (CH₄) isRead the answer and worked solution →Which molecule is non-polar despite having polar bonds?Read the answer and worked solution →The bond order of the oxygen molecule O₂ (by molecular orbital theory) isRead the answer and worked solution →A coordinate (dative) bond differs from a normal covalent bond in thatRead the answer and worked solution →Which species has sp hybridisation on the central atom?Read the answer and worked solution →As bond order increases, bond lengthRead the answer and worked solution →An ionic bond forms byRead the answer and worked solution →The shape of NH₃ is pyramidal rather than planar becauseRead the answer and worked solution →Which has the highest dipole moment?Read the answer and worked solution →Among the following, which is paramagnetic?Read the answer and worked solution →Formal charge on an atom =Read the answer and worked solution →Order the atomic models as they were proposedRead the answer and worked solution →Match each scientist to their modelRead the answer and worked solution →Match each model to what it could NOT explainRead the answer and worked solution →True or false: Rutherford's modelRead the answer and worked solution →

Worked practice questions

For water, K_b = 0.52 K kg mol⁻¹. What is the boiling point of a solution of 18 g glucose (M = 180) in 1 kg water?

  1. 373.20 K
  2. 373.67 K
  3. 374.71 K
  4. 373.15 K
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Correct answer: A. 373.20 K

We have m = (18/180)/1. So m = 0.1 mol kg⁻¹. ΔT_b = 0.52 × 0.1 = **0.052 K**. New b.p. = 373.15 + 0.052 = **373.20 K**.

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Depression of freezing point (ΔT_f) is given by:

  1. ΔT_f = 1000 × m
  2. ΔT_f = K_f × m
  3. ΔT_f = K_b × m
  4. ΔT_f = K_f / m
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Correct answer: B. ΔT_f = K_f × m

**ΔT_f = K_f × m**. K_f is the cryoscopic constant (molal freezing-point depression constant) for water K_f = 1.86 K kg mol⁻¹.

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45 g of ethylene glycol (M = 62) is dissolved in 600 g of water. K_f(water) = 1.86 K kg mol⁻¹. The depression in freezing point is approximately:

  1. 1.5 K
  2. 2.2 K
  3. 0.5 K
  4. 3.5 K
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Correct answer: B. 2.2 K

Moles glycol = 45/62 ≈ 0.73 mol. Mass water = 0.60 kg. m = 0.73/0.60 ≈ 1.21 mol kg⁻¹. ΔT_f = 1.86 × 1.21 ≈ **2.2 K**.

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