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Table of cryoscopic and ebullioscopic constants of substances
Table shows cryoscopic and ebullioscopic constant of selected substances.

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liquids

SubstanceMolecular formulaCryoscopic constant
[kg × K / mol]
Ebullioscopic constant
[kg × K / mol]
water 25°CH2O1.860.513
cyclohexaneC6H1220.82.92
cyclohexanolC6H11OH42.2-
ethylene glycolC2H4(OH)23.112.26
formic acidHCOOH2.38-
acetate acidCH3COOH3.633.22
phenylamine (aniline)C6H5NH25.233.82
benzeneC6H65.072.64
tolueneC6H5CH33.553.4
orthoxyleneC6H4(CH3)25.92-
metaxyleneC6H4(CH3)27.76-
paraxyleneC6H4(CH3)27.2-
chloroformCHCl34.93.8
tetrachloromethaneCCl429.85.3
pyridineC5H5N4.262.83
diethyl ethere(C2H5)2O1.792.02
nitrobenzeneC6H5NO26.875.2

hydrocarbons

SubstanceMolecular formulaCryoscopic constant
[kg × K / mol]
Ebullioscopic constant
[kg × K / mol]
cyclohexaneC6H1220.82.92
benzeneC6H65.072.64
tolueneC6H5CH33.553.4
orthoxyleneC6H4(CH3)25.92-
metaxyleneC6H4(CH3)27.76-
paraxyleneC6H4(CH3)27.2-

alcohols

SubstanceMolecular formulaCryoscopic constant
[kg × K / mol]
Ebullioscopic constant
[kg × K / mol]
cyclohexanolC6H11OH42.2-
ethylene glycolC2H4(OH)23.112.26

carboxylic acids

SubstanceMolecular formulaCryoscopic constant
[kg × K / mol]
Ebullioscopic constant
[kg × K / mol]
formic acidHCOOH2.38-
acetate acidCH3COOH3.633.22

amines

SubstanceMolecular formulaCryoscopic constant
[kg × K / mol]
Ebullioscopic constant
[kg × K / mol]
phenylamine (aniline)C6H5NH25.233.82

other organic

SubstanceMolecular formulaCryoscopic constant
[kg × K / mol]
Ebullioscopic constant
[kg × K / mol]
chloroformCHCl34.93.8
tetrachloromethaneCCl429.85.3
pyridineC5H5N4.262.83
diethyl ethere(C2H5)2O1.792.02
nitrobenzeneC6H5NO26.875.2

inorganic acids

SubstanceMolecular formulaCryoscopic constant
[kg × K / mol]
Ebullioscopic constant
[kg × K / mol]
sulfurous (VI) acidH2SO46.175.33

Some facts

  • The cryoscopic constant tells us how many degrees freezing temperature of 1 kilogram of pure solvent will drop after introducing 1 mole of substance into it.
  • Similarly, the ebullioscopic constant tells us how many degrees boiling point of 1 kilogram of pure solvent will raise after introducing 1 mole of substance into it.
  • Both constants (ebullioscopic and cryscopic) are specific for given solvent.
  • Basic unit for ebullioscopic and cryoscopic constants is kilogram times kelvin per mol:
    kg×Kmol\frac{kg \times K}{mol}
  • In case of so-called perfect solutuion the relationship between the changes of boiling point, the ebullioscopic constant and the composition of the solution is as follows:
    ΔTb=Kb×mM×ms\Delta T_b = K_b \times \frac{m}{M \times m_s}
    where:
    • ΔTb\Delta T_b - changes of boling temperature comparing to pure solvent,
    • KbK_b - ebullioscopic constant for given solvent,
    • mm - mass of solute,
    • MM - molar mass of solute,
    • msm_s - mass of solvent.
  • Analogous, for cryoscopic and freezing point we can write below formula:
    ΔTf=Kf×mM×ms\Delta T_f = K_f \times \frac{m}{M \times m_s}
    gdzie:
    • ΔTk\Delta T_k - changes of freezing point comparing to pure solvent,
    • KkK_k - cryoscopic constant for given solvent,
    • mm - mass of solute,
    • MM - molar mass of solute,
    • msm_s - mass of solvent.

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