### Richard A. Paselk

General Chemistry

Spring 2011

Exercise: Colligative Properties of Solutions

# Colligative Properties Problems

Remember that Colligative properties depend only on the number or concentration of particles in a solution. The properties are, for ideal solutions, independent of the kind or size of the particles, whether ionic, large or small etc. As a consequence:

• For ionic substances the colligative properties depend on the concentration of ions, e.g. for 1M NaCl we have 2M ions = 2M particles, for 1M AlCl3 have 4M ions = 4M.
• For all substances the colligative properties are independent of size or molecular weight, e.g. a 0.01mM solution of ribonuclease (MW = 13,700 daltons) will have the same colligative properties (bp, fp, osmotic pressure, etc.) as a 0.01mM solution of methanol (CH3OH, MW = 32).

In these exercises we will be using the equations for the various colligative properties we have studied:

• Vapor pressure lowering (Raoult's Law): P = XP°, where P° = the vapor pressure of the pure substance and X = its mole fraction.
• Boiling point elevation: Tb = K bm, where m = molality and Kb is a proportionality constant, the molal boiling-point elevation constant specific to the solvent.
• Freezing point depression: Tf = -Kfm, where m = molality and Kf is a proportionality constant, the molal freezing-point depression constant specific to the solvent.
• Osmotic pressure (): V = nRT; or, dividing both sides by V, = MRT, where M = molarity.

Recall also the definitions of the three concentration measures fundamental to solution colligative behaviors:

• X (mole fraction) = moles solute divided by the total number of moles of solutes and solvent.
• For example if we mixed 1.0 moles of methanol + 1.0 moles of ethanol + 3.0 moles of ethylene glycol + 5.0 moles of water the mole fraction of methanol would be, X = 1mol/(1mol + 1mol + 3mol + 5mol) = 1mol/10mol = 0.10.
• Note that X = 1, e.g. X1 + X2 + X3 = 1.
• Note that this is the only measure we are using where ALL solution components appear explicitly.
• m (molality) = moles of solute per kilogram of solvent.
• For example a 1.0m solution of methanol would be made by adding 1.0 mole of methanol (32.0g) to 1.000 kg of water to give 1.032 kg total.
• If there are no other components this is easily expressed as mole fraction. Thus 1.000 x 103g/18.02g/mol = 55.51mol water, so X = 1.0/56.51 = 0.018.
• M (molarity) = moles of solute per liter of solution.
• For example a 1.0M solution of methanol might be made by adding 1.0 mole of methanol (32.0g) to a one L volumetric flask and then adding enough water to bring the total volume to 1.000L.

1. Find the vapor pressure of acetone (CH3COCH3) in a solution made by dissolving 1.00g of a non-volatile compound (sulfanilamide, MW = 172.1) in 10.0g of acetone. The vapor pressure of pure acetone under these conditions is 4.00 x 103mmHg.

2. Determine the total vapor pressure over a solution of 0.600 moles of toluene and 0.400 moles of benzene @ 60 °C assuming ideal behavior. The vapor pressures of pure toluene and benzene are, respectively, 139 mmHg and 392 mmHg @ 60°C.

3. Find the boiling point of a solution of 0.300 g urea (NH2CONH2, MW = 60.1) dissolved in 10.0 g of pure water at 1.00 atm. (Kb = 0.512 °C m-1)

4. Calculate the molecular weight of an unknown substance if dissolving 7.39 g in 85.0 g of benzene (a non-polar solvent) raises the boiling point from 80.2 °C to 82.6 °C. (Kb = 2.52 °C m-1)

5. Calculate the boiling point of a 0.250 m aqueous solution of iron(III) chloride at 1.00 atm. (Kb = 0.512 °C m-1)

6. Calculate the molecular weight of an unknown substance if dissolving 1.42 g in 25.0 g of pure benzene lowers the freezing point by 1.96°C. (Kf = 5.12 °C m-1)

7. What is the freezing point of a 0.100 m aqueous solution of aluminum chloride? (Kf = 1.855 °C m-1)

8. A 0.150 m solution of acetic acid in water is found to have a freezing point of - 0.28 °C. What is the concentration of hydrogen ions in this weak acid solution? (Kf = 1.855 °C m-1)

9. Find the osmotic pressure of a 0.500 M solution of sodium chloride at 0 °C.

10. In order to find the molecular weight of hemoglobin 0.500 g was dissolved in enough water in a volumetric flask to give 100.0 mL of solution. The osmotic pressure of this solution was then measured at 25°C and found to be 1.35 mmHg. Calculate the moleculaar weight.