A 0.561 m soluation of an unknown electrolyte depresses the freezing point of water by 2.93 C . whatis the van't hoff factor for this electrolyte.

Asked by Sunny Acharjee | 13th Jan, 2013, 11:44: AM

Expert Answer:

?TF = KF · b · i

  • ?TF, 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).
?TF = KF · b · i
 2.93 C = (i) (1.86 kg/mol) (0.561 m)

i = 2.8

That's a reasonable value for something like CaCl2. Due to ion-pairing, the measured value for the van 't Hoff factor is less than the predicted value (which for CaCl2 would be 3).

Answered by  | 13th Jan, 2013, 06:11: PM

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