Remember—Colligative properties are properties which depend only on the number or concentration, not on the type, of particles.
Tf = -kfm, where m = molality = moles solute/kg solvent, and kf is a constant specific to the solvent.
Which of the following solutions will have the lowest melting point: 2 m sugar (sucrose) or 0.8 m calcium chloride?
First need to look at concentration of particles.
sugar is covalent, so 2 m particles,
calcium chloride is ionic with 1 mole calcium ions and 2 moles of chloride ions for each mole of CaCl2 = (0.8m)(3mol ions/mol formula) = 2.4 m particles,
Therefore the 0.8 m calcium chloride solution will have the lower mp.
What is the freezing point of the calcium chloride solution?
Freezing point = Fppure solvent – Tf ; kf H2O = 1.855°C/molal
Tf = -kfm = –(1.855°C/m)(2.4m) = –4.452°C
Freezing point = 0.00°C – 4.452°C = –4.4452°C = –4°C
V = nRT; or, dividing both sides by V, = MRT, where M = molarity.
Example: What are the osmotic pressures of 1.00 M sugar and 1 M aluminum chloride solutions at 25°C?
sugar= MRT = (1 mol/L)(0.0821 L*atm/mol*K)(298 K) = 24.5 atm
AlCl3= MRT = (1 mol/L)(4 mol ion/mol)(0.0821 L*atm/mol*K)(298 K) = 97.9 atm.
Liquids: The particles of a liquid are in continuous motion, but the distances between collisions are very short compared to those of gases. Thus liquids are largely incompressible - need to increase pressure about a million-fold to halve volume. Diffusion though liquids is much slower than in gases (hours to days vs. seconds to minutes under standard conditions).
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© R A Paselk
Last modified 27 April 2015