Recently we learnt that there is a depression in the freezing point when a non-volatile solute is added to the solvent. Our teacher explained this phenomenon with a graph where the vapour pressure of the solution and pure solvent intersects with the vapour pressure of the solid solvent. my question is: wouldn't the vapour pressure of the solid solution be lower than the vapour pressure of the solid solvent(that would make another curve in the graph)? Because, the non-volatile solute is still blocking the surface of the solvent, thus reducing the vapour pressure. The solution is of a solid non-volatile solute in a liquid solvent enter image description here

  • $\begingroup$ Blocking the surface is not a part of the picture at all. $\endgroup$ Mar 20 '21 at 10:19

In chemistry we assume a mixture of two solids as two different phases, which do not interfere with each other and thus, the solid solute in the solid mixture will not interfere with the solid solvent. So, the vapour pressure of the solid solution and solid solvent will always be same. In a similar way, when two immiscible liquids are mixed together, the total vapour pressure is counted as the sum of Vapour pressures of the liquids if kept separately.

Hope this helps!!

  • 2
    $\begingroup$ Unless the compounds form solid solutions, if they have similar enough properties. But that is rarely the case for the volatile and non-volatile components. $\endgroup$
    – Poutnik
    Mar 20 '21 at 8:51
  • $\begingroup$ @Poutnik Thanks for adding the value!! $\endgroup$ Mar 20 '21 at 9:09

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