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Colligative Properties and calculations involving colligative properties

Submitted by mariana582 on Sun, 04/07/2013 - 17:52

Hi,  i need help with colligative properties especially mole fraction and freezing/boiling point depression. Sometimes I am able to master these properties but most of the the time I don't. I really need to master this concept because I am still not quite comfortable with them. With most problems I don't know were to begin even though I have all the formulas in front of me. For example,


What mass of water is needed to dissolve 34.8 g of copper(II) sulfate in order 
 to prepare a 0.521 m solution? 
Calculate the freezing point of a solution made from 32.7 g of propane, 
C3H8, dissolved in 137.0 g of benzene, C6H6. The freezing point of benzene 
 is 5.50° C and its Kf
 is 5.12° C/m.\
Calculate the concentration of nitrogen gas in a 1.00 L container exerting a 
 partial pressure of 572 mm Hg at room temperature. Henry’s law constant 
 for nitrogen at 25° C is 6.8 x 10-4 mol/L·atm.
Thank You!

Find molar mass of CuSO4 or CuSO4.5H2O if that's what you used. Moles = 34.8/molar mass. Then 0.521 moles are in 1 kg of solvent so (34.8/molar mass) moles will be in (1/0.521) x (34.8/molar mass) kg of solvent.

2) find moles propane from mass/molar mass. Molality = moles per kg solvent.

From Wikipedia

ΔTF = KF · b · i

  • ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF (solution).
  • KF, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. Note: When conducting experiments, a higher KF value makes it easier to observe larger drops in the freezing point. For water, KF = 1.853 K·kg/mol.[4]
  • b is the molality (mol solute per kg of solvent)
  • i is the van 't Hoff factor (number of ion particles per individual molecule of solute, e.g. i = 2 for NaCl, 3 for BaCl2)