Based on the comment to an answer elsewhere by OP, I'd try to solve this problem, purely based on volumes. However, these calculations has ignored the volume contraction of 96% ethanol may have showed initially (that should be minimal since it is only ~4% of water by volume in there). I'd say, your equation, regardless of how OP has derived it, is erroneous. I also like to change the variables OP has used to followings:
Accordingly, the needed volume of pure ethanol $ = V_{Tot} \times \frac{P_\%}{100}=0.96V_{E}$. Thus, $$ V_{E} = \frac{V_{Tot} \times P_\%}{100 \times 0.96}= \frac{V_{Tot} \times P_\%}{96} \tag{1}$$ Note that this is actually OP's first equation, but it is for $V_{E}$ ($E$ in OP's notation) instead of $V_{W}$ ($W$ in OP's notation). Now we can derive the equation for $V_{W}$. Actual $V_{W}$ is: $$V_{W}=V_{Tot}\left(\frac{100-P_\%}{100}\right)$$ Yet, we cannot use $V_{W}=V_{Tot}-V_{E}$, because some water is coming from $V_{E}$. That amount of water is $ 0.04V_{E} = 0.04 \times \frac{V_{Tot} \times P_\%}{96}$. Thus, we can manipulate this equation as follows: $$V_{W}=(V_{Tot}-V_{E})-0.04 \times \frac{V_{Tot} \times P_\%}{96}= V_{Tot}\left(\frac{100-P_\%}{100}\right)-0.04 \times \frac{V_{Tot} \times P_\%}{96}\\=\frac{96-P_\%}{96}\times V_{Tot} $$ $$\therefore \; V_{W}=\frac{96-P_\%}{96}\times V_{Tot} \tag{2}$$ Now we apply these two equation to OP's example of making 50% solution: If $P_\% = 50$ and $V_{Tot} = \pu{50 L}$, from equation $(1)$ and $(2)$, $$ V_{E} = \frac{V_{Tot} \times P_\%}{96}= \frac{50 \times 50}{96} = \pu{26.042 L}$$ $$V_{W}=\frac{96-P_\%}{96}\times V_{Tot} = \frac{96-50}{96}\times 50 = \pu{23.958 L}$$ Thus, theoretical total (disregarding contraction) is $\pu{50 L}$. Acutally, practical volume must be a little off but your percentage by volume is much close to 50% (only volume contraction did not account is initial 96% ethanol). Late edition to fulfill OP's request: If $P_\% = 35$ and $V_{Tot} = \pu{100 L}$, from equation $(1)$ and $(2)$, $$ V_{E} = \frac{V_{Tot} \times P_\%}{96}= \frac{100 \times 35}{96} = \pu{36.458 L}$$ $$V_{W}=\frac{96-P_\%}{96}\times V_{Tot} = \frac{96-35}{96}\times 100 = \pu{63.542 L}$$ Question
Pour 200 milliliters of alcohol into a graduated cylinder. Then, pour 200 milliliters of water Identify the liquids as alcohol and water. Observe traces of liquid inside the contains first holding the liquids. Recognize that the traces of liquid could not account for the loss in volume. When I actually performed this demo in front of my colleagues I added food coloring to both the water and lacohol to show them mixing more clearly and I used 95% ethanol and their was know appreciable loss of volume. It was suggested to use rubbing alcohol (isopropyl alcohol) instead at the highest concentration you can find. Perhaps the food coloring molecules interferred with the water molecules from moving into the interstitial spaces between the larger alcohol chains but then again ethanol does not react the expected way as mentioned in the sources I found. Sorry all for failing to make the magic happen.
Bill Nye Science Guy showing the discrepant event of water and alcohol. In the case of ethanol (alcohol) and water the volume of some concentrations is less than the sum of the components. Liquid water has a somewhat "open" structure that is broken up by the addition of ethanol so the mixture "collapses". The emission spectra obtained in the study of methanol and water reveal that the water and alcohol molecules in solution form complex hydrogen-bonded networks and mix very little at the microscopic level. The results illustrate the technique's potential to provide new and valuable information about the microscopic origins of the properties of liquids and solutions. The results show that the structure of liquid methanol at room temperature is a combination of rings and chains, each made up of either 6 or 8 methanol molecules. When water is added, the methanol chains interact with varying numbers of water molecules. These "bridging" water molecules bend the chains into open-ring structures that are stable because their glue-like hydrogen bonds are saturated. This means that the mixing of alcohol and water on the microscopic level is incomplete no matter how long you wait.
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