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Colligative Properties. Honors Chemistry Unit 8 Chapter 15. Colligative Properties. These are the effects that a solute has on a solvent. When water has something dissolved in it, its physical properties change.
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Colligative Properties Honors Chemistry Unit 8 Chapter 15
Colligative Properties • These are the effects that a solute has on a solvent. • When water has something dissolved in it, its physical properties change. • It will no longer boil at 100oC and it will no longer freeze at 0oC like pure water.
Three Main Effects • Lowers the vapor pressure of a solvent. • Lower vapor pressure means that fewer water molecules can escape from the liquid phase into the gas phase at given temperature. Remember, a lower vapor pressure means a higher boiling point! • 2) Raises the boiling point of a solvent. • 3) Lowers the freezing point of a solvent.
Applications • salting icy roads • making ice cream • antifreeze • cars (-64°C to 136°C) • fish & insects
Colligative Properties -These depend only on the numberof dissolved particles -Not on what kind of particle General Rule: The more solute particles that are present in a solvent, the greater the effect.
# of Particles • Nonelectrolytes (covalent) • remain intact when dissolved • 1 particle • Electrolytes (ionic) • dissociate into ions when dissolved • 2 or more particles • Electrolytes have a stronger affect in lowering the freezing point and elevating the boiling point because it puts more particles into the solution.
Remember the rule, the more particles, the greater the effect! • The Dissociation Factor (d.f.) for an electrolyte is the number of ions a compound dissociates into. • NaCl gives Na+ ions and Cl- ions, which is 2 particles, therefore d.f.= 2 • What is “d.f.“ for Al(NO3)3 ? Al(NO3)3 Al3+ + 3 NO3- = 4 particles
Non-electrolytes • A compound that does not conduct electricity when dissolved in water. • Examples are glucose or any other sugars and alcohols, such as ethanol (CH3CH2OH) • The d.f. = 1 for any non-electrolyte. • Why? • Covalent molecules do not break apart when they become solvated by water molecules.
Calculations t = m · d.f. · K t: change in temperature (°C) K: constant based on the solvent (°C·kg/mol) Use Kb for boiling point elevation and Kf for freezing point depression. Each solvent has it’s own unique factors! m: molality (m) = moles solute/Kg solvent d.f.: # of particles
Example 1: If 48 moles of ethylene glycol, C2H4(OH)2, is dissolved in 5.0kg of water. What is the boiling point of the solution? • ∆Tbp = m d.f. Kb • d.f. = 1 ( it is a non-eletrolyte) • Kb = 0.52oC/m (a constant for water) • Molality = m = moles solute/kg solvent ∆Tbp = (48 moles) (1) (0.52oC/molal) (5.0 Kg) New BP = 100 + 5 = ∆Tbp = 5.0 oC 105 oC
Example 2: 55.5 grams of CaCl2 are dissolved in 2.0kg of water. What is the freezing point of this solution? • ∆Tfp = m d.f. Kf • d.f. = 3 (CaCl2 produces 3 particles) • CaCl2 Ca2+ + 2 Cl- • Kf = 1.86oC/m (a constant for water) Must calculate molality Molality = mass/molar mass Kg Solvent = (55.5 gram/111g/mole) 2.0 kg = 0.25m ∆Tfp = (0.25molal) (3) (1.86oC/molal) = 1.4 oC New FP = -1.4 oC