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I am conducting an experiment where I mix two solutions $\text{A}$ and $\text{B}$ and then I am measuring the temperature change (for real solutions, i.e., $\Delta_\mathrm{mix}H\neq 0).$ I took the following mixtures which showed the following changes in temperature:

$$ \begin{array}{lllcc} \hline \# & \text{A} & \text{B} & \Delta T_\mathrm{surr} & \Delta_\mathrm{mix}H \\ \hline 1 & \text{Acetone} & \text{Aniline} & > 0 & < 0 \\ 2 & \text{Acetone} & \text{Chloroform} & > 0 & < 0 \\ 3 & \text{Methanol} & \text{Acetic acid} & > 0 & < 0 \\ 4 & \text{Acetone} & \text{Ethanol} & < 0 & > 0 \\ 5 & \text{Ethanol} & \text{Water} & > 0 & \color{red}{< 0} \\ \hline \end{array} $$

The above mentioned are the results of my experiment which I conducted in my school chemistry lab at temperatures around $\pu{30 °C}$ with approximately no change in atmospheric pressure.

As seen, the first four cases show results as expected. However, in the 5th case, it's written all over the internet that an ethanol-water mix shows positive deviation, however, as seen through my experimental results, $\Delta_\mathrm{mix}H < 0$, i.e., it shows negative deviation. Can someone please tell me why this is happening?

Note: I repetitively cleaned all my equipment thoroughly with water before using the apparatus. Also, I conducted 3 trials in total for each mix and they showed similar results, so there is no chance of random errors leading to outliers.

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    $\begingroup$ They must mean positive deviation from Raoult law, with mixture having higher vapour tension than expected? As it is well known ethanol-water mixing enthalpy is negative. $\endgroup$
    – Poutnik
    Oct 25 '21 at 8:09
  • $\begingroup$ @Poutnik Yes, it is a positive deviation from Raoult's law, but wouldn't that also mean that the temperature would fall for the surrounding since the enthalpy of intramolecular bond formation after the solutions are mixed is greater than the enthalpy of intramolecular bond formation for pure solutions? $\endgroup$ Oct 25 '21 at 8:11
  • $\begingroup$ The enthalpy is lower, ergo the mixing is exothermic $\endgroup$
    – Alchimista
    Oct 25 '21 at 9:07