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Chemistry 3% exam weight

Forces Between Particles

Part of the UTBK/SNPMTN (Indonesia) study roadmap. Chemistry topic chemis-010 of Chemistry.

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Forces Between Particles

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Intermolecular forces (IMFs) hold molecules together in the liquid and solid phases and dictate boiling point, melting point, and solubility. Strength ranking: ion–dipole > hydrogen bonding > dipole–dipole > London dispersion.

“Like dissolves like” — polar/ionic solutes match polar solvents; non-polar solutes match non-polar solvents.

Colligative properties depend on particle count, not identity. The four key equations:

PropertyEquationUnits of constant
Boiling-point elevationΔTb = Kb · m · iKb in °C·kg/mol
Freezing-point depressionΔTf = Kf · m · iKf in °C·kg/mol
Osmotic pressureπ = i · M · R · TR = 0.0821 L·atm/(mol·K)
Vapor-pressure loweringΔP = (1 − x_solvent) · P°Raoult’s law

Use molality (mol/kg solvent) for ΔTb and ΔTf; use molarity (mol/L solution) for π. Apply van’t Hoff factor i: i ≈ 2 for NaCl, i ≈ 3 for CaCl₂.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Forces Between Particles

IMFs differ from covalent bonds; they act between molecules. Hydrogen bonding requires H bonded directly to N, O, or F — a frequent UTBK trap, since any molecule with N–H or O–H doesn’t automatically qualify.

  • Ion–dipole: ion + polar molecule (e.g., Na⁺ with H₂O); strongest of the four.
  • Hydrogen bonding: special, strong dipole–dipole case in H₂O, HF, NH₃.
  • Dipole–dipole: between permanent polar molecules.
  • London dispersion: instantaneous dipoles in all molecules; dominant in non-polar species and increases with molar mass and surface area.

Stronger IMFs → higher boiling and melting points, lower vapor pressure, lower solubility of gases.

Concentration Units

UnitDefinitionUsed in
Molarity (M)mol solute / L solutionπ equation
Molality (m)mol solute / kg solventΔTb, ΔTf
Mole fraction (x)mol component / total molRaoult’s law
Mass percent(mass solute / mass solution) × 100%General composition

Mixing molarity with molality is the most common UTBK error. ΔTb and ΔTf need molality because they depend on temperature-invariant solvent mass.

Colligative Properties in Practice

ΔTb raises the boiling point; ΔTf lowers the freezing point. Solutes also lower vapor pressure (Raoult’s law) and generate osmotic pressure π. For electrolytes, multiply by the van’t Hoff factor i to count dissociated particles: CaCl₂ → i ≈ 3, K₃PO₄ → i ≈ 4. Non-electrolytes keep i = 1.

A typical UTBK item: “How many grams of NaCl (M = 58.5 g/mol) in 500 g water give ΔTf = 3.72 °C, Kf_water = 1.86 °C·kg/mol?” Solve m = ΔTf/(Kf·i) = 3.72/(1.86·2) = 1 mol/kg → 0.5 mol → 29.25 g.

Exam Pointers

  • UTBK Kimia gives Topic 10 roughly 1–3 questions (~3% weight), usually within the Pemahaman Konsep dan Aplikasi Kimia sub-test.
  • Common formats: rank molecules by boiling point, identify the strongest IMF, calculate ΔTb/π given mass and Kb.
  • Watch unit conversions: °C → K when using π = iMRT; kg ↔ g for molality; atm ↔ Pa for pressure.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Real Behaviour

The “ideal” van’t Hoff factor (NaCl → 2, CaCl₂ → 3) is a textbook maximum. At high concentrations, ion pairing reduces the effective i below the theoretical value — so experimentally measured ΔTb is smaller than predicted. UTBK rarely tests this directly, but the principle explains why real solutions deviate from linear colligative plots.

For volatile solutes (e.g., benzene + toluene), Raoult’s law becomes P_total = x₁P°₁ + x₂P°₂. Applying the single-component form ΔP = (1 − x_solvent)P°_solvent to such a mixture underestimates total vapor pressure and is a classic trap.

Worked Example — Osmotic Pressure

A 2.00 L solution contains 5.85 g NaCl at 27 °C. Find π in atm.

Step 1 — moles: 5.85 g ÷ 58.5 g/mol = 0.100 mol NaCl. Step 2 — molarity: 0.100 mol ÷ 2.00 L = 0.0500 mol/L. Step 3 — i for NaCl ≈ 2, T = 300 K. Step 4 — π = iMRT = 2 × 0.0500 × 0.0821 × 300 = 2.463 atm.

Reversing the logic: molecular weight of an unknown solute can be found from π = (mass·R·T)/(M·V), a frequent UTBK application.

Connections to Adjacent Topics

IMF strength correlates with enthalpy of vaporization (Topic 6, Thermochemistry) and explains trends in vapor pressure curves on phase diagrams (Topic 5). Solubility rules link to electrolyte equilibria (Topic 9) via the same dissociation logic that sets i.

Common Mistakes

  • Forgetting i → underestimating ΔTf/ΔTb/π by 2× or 3×.
  • Using molarity in ΔTf → wrong answer whenever temperature changes solvent volume.
  • Treating any N–H or O–H bond as hydrogen bonding (requires N, O, or F on H).
  • Equating ΔTb direction with ΔTf direction (one raises the curve, the other lowers it).

Practice Prompts

  1. Rank by boiling point: CH₄, H₂O, HF, NH₃ — justify using IMF type and molar mass.
  2. A solution of 6.00 g urea (M = 60 g/mol, i = 1) in 250 g water: compute ΔTb (Kb_water = 0.52 °C·kg/mol) and predict the new boiling point.

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