How to calculate the freezing point of an equimolar ethanol: water mixture using colligative properties [closed]








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I am familiar with how to calculate the freezing point of a mixture with a 9:1 molar ratio of ethanol to water (see below). I can understand that ethanol, having a higher mole fraction, is considered to be the solvent. Hence in this case, we are finding the depression in freezing point assuming water to be the solute.


If I were to modify the question so that equal amounts of ethanol and water are present in the mixture, how would one proceed to find the depression in freezing point/elevation in freezing point of the mixture? What is the solvent and what is the solute in that case?


[![enter image description here](https://i.stack.imgur.com/BBsES.jpg)](https://i.stack.imgur.com/BBsES.jpg)


Credits:solution provided by <https://www.doubtnut.com/>





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> What is the solvent and what is the solute in that case?
> 
> 
> 


To explore this, you could try both. You would get two different answers, though.



> 
> how would one proceed to find the depression in freezing point/elevation in freezing point of the mixture?
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> 
> 


Colligative property expressions work well for low concentrations (where the solutes behave like they are "ideally diluted"). If you look at the experimental data below, you realize that for higher concentrations, the change in freezing point is no longer linear:


[![enter image description here](https://i.stack.imgur.com/mEphh.png)](https://i.stack.imgur.com/mEphh.png)


Source: <https://commons.wikimedia.org/wiki/File:Phase_diagram_ethanol_water_s_l_en.svg>


You also see that the 1:9 water: ethanol mixture might be at the edge of where the freezing point depression formula is useful (note the scale in the image is a mass ratio, not an amount ratio).


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