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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:

• $V_{Tot}$ is the desired Total amount of desired ethanol-water solution (in liters);
• $P_\%$ is the desired Percentage of ethanol in the final solution (in $\%(v/v)$;
• $V_{W}$ is the calculated amount of Water (in liters); and
• $V_{E}$ is the calculated amount of $96\%(v/v)$ ethanol (in liters).

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).

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:

• $V_{Tot}$ is the desired Total amount of desired ethanol-water solution (in liters);
• $P_\%$ is the desired Percentage of ethanol in the final solution (in $\%(v/v)$;
• $V_{W}$ is the calculated amount of Water (in liters); and
• $V_{E}$ is the calculated amount of $96\%(v/v)$ ethanol (in liters).

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 $(2)$,

$$ V_{E} = \frac{V_{Tot} \times P_\%}{96}= \frac{100 \times 35}{96} = \pu{36.458 L}$$

$$V_{W}=\frac{96-P_\%}{96}\times V_{Tot} = \frac{96-35}{96}\times 100 = \pu{63.542 L}$$

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:

• $V_{Tot}$ is the desired Total amount of desired ethanol-water solution (in liters);
• $P_\%$ is the desired Percentage of ethanol in the final solution (in $\%(v/v)$;
• $V_{W}$ is the calculated amount of Water (in liters); and
• $V_{E}$ is the calculated amount of $96\%(v/v)$ ethanol (in liters).

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 notification) instead of $V_{W}$ ($W$ in OP's notification).

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).

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:

• $V_{Tot}$ is the desired Total amount of desired ethanol-water solution (in liters);
• $P_\%$ is the desired Percentage of ethanol in the final solution (in $\%(v/v)$;
• $V_{W}$ is the calculated amount of Water (in liters); and
• $V_{E}$ is the calculated amount of $96\%(v/v)$ ethanol (in liters).

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).

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:

• $V_{Tot}$ is the desired Total amount of desired ethanol-water solution (in liters);
• $P_\%$ is the desired Percentage of ethanol in the final solution (in $\%(v/v)$;
• $V_{W}$ is the calculated amount of Water (in liters); and
• $V_{E}$ is the calculated amount of $96\%(v/v)$ ethanol (in liters).

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 notification) instead of $V_{W}$ ($W$ in OP's notification).

Now we can derive

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:

• $V_{Tot}$ is the desired Total amount of desired ethanol-water solution (in liters);
• $P_\%$ is the desired Percentage of ethanol in the final solution (in $\%(v/v)$;
• $V_{W}$ is the calculated amount of Water (in liters); and
• $V_{E}$ is the calculated amount of $96\%(v/v)$ ethanol (in liters).

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 notification) instead of $V_{W}$ ($W$ in OP's notification).

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).