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Chemistry · Ch 9 — Solutions

Depression in Freezing Point

9.9.3

Depression in Freezing Point

Freezing point is defined as the temperature at which the solid and liquid states of a substance have the same vapour pressure -- i.e. the solid and liquid phases are in equilibrium. Water's freezing point, for example, is 0∘0^\circC, the temperature at which ice and liquid water coexist in equilibrium.

When a nonvolatile solute is added to water at its freezing point, the freezing point of the resulting solution is lower than 0∘0^\circC. This lowering, when a solute is added to a solvent, is called depression in freezing point, ΔTf\Delta T_f:

ΔTf=Tf∘−Tf\Delta T_f = T_f^\circ - T_f

Reading the vapour-pressure-versus-temperature picture (Figure 9.12): the pure solvent's freezing point Tf∘T_f^\circ is where the solid-solvent curve and the pure-liquid-solvent curve cross (at 11 atm, 0∘0^\circC for water). The solution's liquid curve sits below the pure solvent's throughout (lower vapour pressure, exactly as in section 9.9.2), so it crosses the (unchanged) solid curve at a lower temperature Tf<Tf∘T_f<T_f^\circ -- this is the solution's freezing point.

Experimentally, this depression is directly proportional to the molal concentration of solute:

ΔTf∝m\Delta T_f \propto m

ΔTf=Kf m(9.25)\Delta T_f = K_f\,m \qquad (9.25)

where KfK_f is the molal freezing-point-depression constant (also called the cryoscopic constant) of the solvent. As with KbK_b, if m=1m=1, then ΔTf=Kf\Delta T_f=K_f -- the depression produced by exactly one mole of solute per kilogram of solvent.

Table 9.4 -- KfK_f values for common solvents (K kg mol−1^{-1}): water 1.86; ethanol 1.99; benzene 5.12; chloroform 4.79; carbon disulphide 3.83; ether 1.79; cyclohexane 20.0; acetic acid 3.90.

Determining molar mass from freezing point depression. For a solution of wBw_B g solute in wAw_A g solvent, molality is

m=wB/MBwA×1000(9.26−9.28)m = \frac{w_B/M_B}{w_A}\times1000 \qquad (9.26-9.28)

so

ΔTf=Kf wBMB wA×1000(9.29)\Delta T_f = \frac{K_f\,w_B}{M_B\,w_A}\times1000 \qquad (9.29)

which rearranges to give the solute's molar mass:

MB=Kf wBΔTf wA×1000(9.30)M_B = \frac{K_f\,w_B}{\Delta T_f\,w_A}\times1000 \qquad (9.30) …

Figure 9.12Depression in freezing point

What this figure shows. Temperature (x-axis) plotted against vapour pressure (y-axis) at atmospheric pressure (1 atm). The solid-solvent (frozen) curve and the pure-solvent liquid curve cross at Tf∘T_f^\circ (0∘^\circC for water). A second liquid curve, for the solution, sits below the pure solvent's and crosses the solid curve at a lower temperature TfT_f, with the horizontal gap ΔTf=Tf∘−Tf\Delta T_f=T_f^\circ-T_f mark …

Table 9.4Molal freezing point depression constant Kf for some solvents
SolventFreezing point (K)Kf_f (K kg mol−1^{-1})
Water273.01.86
Ethanol155.71.99
Benzene278.65.12
Chloroform209.64.79
Carbon disulphide164.23.83
Ether156.91.79
Misc Example Problem 5Freezing point of a 20% glycol antifreeze mixture

Worked out. Ethylene glycol (C2_2H6_6O2_2, molar mass 62 g mol−1^{-1}) is used as radiator antifreeze; find when ice begins to separate from a mixture that is 20 mass percent glycol in water (Kf=1.86K_f=1.86 K kg mol−1^{-1} for water). Weight of solute w2=20w_2=20 g, weight of solvent w1=100−20=80w_1=100-20=80 g. ΔTf=Kfm=Kf×w2×1000M2×w1=1.86×20×100062×80=7.5\Delta T_f=K_fm=\dfrac{K_f\times w_2\times1000}{M_2\times w_1}=\dfrac{1.86\times20\times1000}{62\times80}=7.5 K. Ice begins to separate 7.5 K below water's normal freezing point, i.e. at $27 …

Misc Evaluate Yourself 12Molar mass of a solute from freezing point depression in benzene

Worked out. An in-text practice box: 2 g of a non-electrolyte solute dissolved in 75 g of benzene lowers benzene's freezing point by 0.20 K (Kf=5.12K_f=5.12 K kg mol−1^{-1} for benzene). Find the molar mass of the solute. …