It is known that addition of a non-volatile solute to a volatile solvent(liquid) to give a solution reduces the Vapour Pressure of the solution (well solvent actually as only solvent is volatile).

This leads to the Elevation of Boiling Point and "Deppresion" of Freezing Point.

I am clear with elevation of BP. But it is said that a liquid freezes when the vapour pressure of the liquid phase attains the vapour pressure of its solid state.

So, how possibly does the freezing point falls? For instance let the Vapour Pressure of the solid state be at an arbitrary point "x"(at a temperature A Kelvin) and that of the liquid phase be at "y".

So when we add a solute to the liquid:

Vapour pressure of liquid(solution) = y - $\alpha $ (where $\alpha $ is the depression in the vapour pressure)

So now the vapour pressure of the liquid(solution) reaches "x" rapidly that is at a higher temperature or a temperature greater than A which is an elevation in the freezing point.

I am always confused with this part.


1 Answer 1


Focus more on free energies rather than on vapor pressures (which derive, ultimately, from free energies after all). For a mixture of B (solute) in A (solvent), the entropy of mixing is $RT(x_A ln(x_A) + x_B ln(x_B))$, and the enthalpy of mixing will go approximately as $x_A x_B \Omega$, with $\Omega$ as a measure of the interaction of A and B. The entropy term will always result in a reduction of free energy at small $x_B$ regardless of the sign of $\Omega$, but in the case of salt in water $\Omega$ is negative, driving further solubility.

Now, about those temperatures of phase transitions (which as you should recall occur when the free energies of the phases are equal, more fundamental than vapor pressures). On the boiling end, water with salt in it has a lower free energy than water without salt - the boiling point of the salt water has to be higher than for pure water. The presence of the solute makes it happy to stay liquid.

So, why isn't the freezing point raised in an analogous way? For pure A, of course, the freezing point remains that of pure water. The entropy of mixing is equivalent for solid and liquid in this case, and it is not clear how different the excess enthalpy is going to be - so why does liquid continue to be stable? Well, it only does while it has a higher concentration of salt in it than is in the solid. It forms a classic eutectic point on the binary phase diagram (see, for example http://antoine.frostburg.edu/chem/senese/101/solutions/images/saltwater-phase-diagram.gif). The only requirement of melting point depression is that water at a high concentration of salt has a lower free energy than the solid at a lower concentration of salt. Given the outline of free energy above, this is likely to hold for some range of temperatures.

Again, remember that what is being lowered is the freezing point of water with salt in it - what starts freezing out is ice with a lower salt content in it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.