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I have trouble understanding as to what the last part of the increase in liquid level means. I believe this is the foam formed after the fermentation? I am calculating for the pressure but I have trouble as to what volume should I use to find the moles since I don't understand the last part. It would greatly help me to understand this process first.

I've been extensively searching how the fermentation process actually occurs cause I don't understand a thing about it. To find the pressure from what I know, I converted the volume of glucose to $\ce{CO2}$ to find the moles and plugged the value in the ideal gas equation $p=nRT/v$ with the volume being $\pu{0.07L}$ from $(\pu{125mL} - \pu{55mL})$, because it says the $\ce{CO2}$ in solution is negligible.

I don't know where the changes in densities come in and I don't know how to incorporate the pressure brought about by the liquid solution. I think my thought process is wrong really cause I don't understand fermentation.

https://www.homebrewtalk.com/threads/calculating-total-gas-produced-in-fermentation.165020/ This site proposed a $p_1v_1=p_2v_2$ method where $p_1$ is at STP, but I don't think this is needed in the problem.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – andselisk Oct 16 at 5:32
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Alcohol fermentation is the formation of alcohol from sugar (or glucose). Yeast, when under anaerobic conditions, convert one glucose molecule $(\ce{C6H12O6})$ to two molecules of pyruvic acid $(\ce{CH3C(=O)CO2H})$ via the glycolysis pathways, then go one step farther, converting each pyruvic acid into ethanol $(\ce{CH3CH2OH})$ and carbon dioxide $(\ce{CO2})$ via acetaldehyde:

Glycolysis process

Thus, each glucose molecule would form two ethanol molecules and two $\ce{CO2}$ molecules:

$$\ce{C6H12O6 ->[yeast] 2C2H5OH + 2 CO2}$$

Let's look at what we have in the flask: $\pu{50 mL}$ of $20\% (w/v)$ glucose solution, density of which is $\pu{1.08 g mL-1}$. Thus, mass of $\pu{50 mL}$ solution is: $$\pu{50 mL} \times \pu{1.08 g/mL} = \pu{54 g}$$

Since $\pu{10 g}$ of this mass is the mass of glucose, the mass of water in the $\pu{50 mL}$ solution is $\pu{44 g}$. Assuming density of water is $\pu{1.0 g mL-1}$ at $\pu{30 ^\circ C}$, then the volume of water at this temperature is $\pu{44 mL}$ (Note: This assumption is justified by the facts that neither densities of solutions are given in particular temperatures nor the $\pu{50 mL}$ of solution is measured in same temperature).

If the solution have gone to 100% completion, none of $\pu{10 g}$ of glucose $(MW = \pu{108.16 g mol-1})$ will be remaining in the solution and as a result, the volume would decrease to $\pu{44 mL}$ of water. Let's calculate the mass ($m$) of ethanol $(MW = \pu{46.07 g mol-1})$ would be produced by this fermentation:

$$m = \frac{\pu{10 g}}{\pu{180.16 g mol-1}} \times 2 \times \pu{46.07 g mol-1} = \pu{5.11 g}$$

Therefore maximum volume of ethanol would produce by fermentation is: $\frac{\pu{5.11 g}}{\pu{0.789 g mL-1}} = \pu{6.48 mL}$. Thus, assuming ethanol and water are additive, maximum volume of the solution in the flask is: $\pu{6.48 mL} + \pu{44 mL} = \pu{50.48 mL}$.

Thus, this reaction would increase the volume of the solution only by $\pu{0.48 mL}$, if the reaction went to 100% conversion, so that the data given is incorrect (can't increase by $\pu{5.00 mL}$ as given).

Thus, assuming 100% conversion, we can calculate the amount of $(\ce{CO2})$ produced ($m_1$) in the flask:

$$m_1 = \frac{\pu{10 g}}{\pu{180.16 g mol-1}} \times 2 = \pu{0.111 mol}$$

Assuming no $(\ce{CO2})$ would dissolve in aqueous solution (this is actually incorrect assumption but it was instructed in the question) and $(\ce{CO2})$ behaves as an ideal gas, the partial pressure inserted by the $(\ce{CO2})$ gas ($p$) is:

$$p = \frac{nRT}{v} = \frac{\pu{0.111 mol} \times \pu{0.0821 atm L mol-1 K-1} \times \pu{303 K}}{\pu{(0.125-0.05048) L}} = \pu{37.1 atm}$$

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    $\begingroup$ I see I was misreading w/v % as w/w % :-) $\endgroup$ – Poutnik Oct 15 at 21:47
  • $\begingroup$ @Poutnik: Do you want to change your answer accordingly? I also didn't understand what $x$-column means. $\endgroup$ – Mathew Mahindaratne Oct 15 at 21:54
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    $\begingroup$ Somewhen in the morning CEDT :-) X is conversion factor, few times mentioned in comments. $\endgroup$ – Poutnik Oct 15 at 21:56
  • $\begingroup$ Thinking by the task author pressurized CO2 would not dissolve is quite funny. Or, combined with the mass issue, it is rather sad, being the task for the task, detached from reality. Interesting would be to know, how [CO2(l)] eventually affect the yeast growth. $\endgroup$ – Poutnik Oct 16 at 6:08
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It seems to me like the task description is wrong, like if they maybe do not consider the decrease of mass of the liquid. As the volume increase due lower density is well compensated by the mass decrease.

If I do not count with decrease of the solution mass, I can get the final volume $\pu{55 mL}$.

Note the $x$ is the glucose conversion factor. $x-0$ means glucose only, $x=1$ means ethanol only.

"V wrong" means the volume change, that is accounting with density decrease due glucose -> ethanol conversions, but not accounting with mass decrease of the liquid.

$\ce{CO2}$ calculations are done by Mathew, no need to doubling that.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{x} & \text{H2O[g]} & \text{glu[g]} & \text{eth[g]} &\text{CO2[g]} & \text{liq[g]} & \text{Eth w/w%} & \text{ $\rho$ g/mL} & \text{ V[mL]} & \text{ V[mL] wrong}\\ \hline 0 & 44.0 & 10.00 & 0.00 & 0.00 & 54.00 & 0.00 & 1.080 & 50.00 & 50.00 \\ \hline 0.05 & 44.0 & 9.50 & 0.26 & 0.24 & 53.76 & 0.48 & 1.075 & 50.01 & 50.24 \\ \hline 0.1 & 44.0 & 9.00 & 0.51 & 0.49 & 53.51 & 0.97 & 1.070 & 50.02 & 50.48 \\ \hline 0.15 & 44.0 & 8.50 & 0.77 & 0.73 & 53.27 & 1.46 & 1.065 & 50.04 & 50.72 \\ \hline 0.2 & 44.0 & 8.00 & 1.02 & 0.98 & 53.02 & 1.96 & 1.059 & 50.05 & 50.97 \\ \hline 0.25 & 44.0 & 7.50 & 1.28 & 1.22 & 52.78 & 2.46 & 1.054 & 50.06 & 51.22 \\ \hline 0.3 & 44.0 & 7.00 & 1.53 & 1.47 & 52.53 & 2.96 & 1.049 & 50.07 & 51.47 \\ \hline 0.35 & 44.0 & 6.50 & 1.79 & 1.71 & 52.29 & 3.47 & 1.044 & 50.09 & 51.72 \\ \hline 0.4 & 44.0 & 6.00 & 2.04 & 1.96 & 52.04 & 3.99 & 1.039 & 50.10 & 51.98 \\ \hline 0.45 & 44.0 & 5.50 & 2.30 & 2.20 & 51.80 & 4.51 & 1.034 & 50.11 & 52.24 \\ \hline 0.5 & 44.0 & 5.00 & 2.56 & 2.44 & 51.56 & 5.04 & 1.029 & 50.12 & 52.50 \\ \hline 0.55 & 44.0 & 4.50 & 2.81 & 2.69 & 51.31 & 5.57 & 1.023 & 50.14 & 52.76 \\ \hline 0.6 & 44.0 & 4.00 & 3.07 & 2.93 & 51.07 & 6.10 & 1.018 & 50.15 & 53.03 \\ \hline 0.65 & 44.0 & 3.50 & 3.32 & 3.18 & 50.82 & 6.64 & 1.013 & 50.16 & 53.30 \\ \hline 0.7 & 44.0 & 3.00 & 3.58 & 3.42 & 50.58 & 7.19 & 1.008 & 50.18 & 53.57 \\ \hline 0.75 & 44.0 & 2.50 & 3.83 & 3.67 & 50.33 & 7.74 & 1.003 & 50.19 & 53.85 \\ \hline 0.8 & 44.0 & 2.00 & 4.09 & 3.91 & 50.09 & 8.30 & 0.998 & 50.20 & 54.12 \\ \hline 0.85 & 44.0 & 1.50 & 4.34 & 4.16 & 49.84 & 8.86 & 0.993 & 50.22 & 54.40 \\ \hline 0.9 & 44.0 & 1.00 & 4.60 & 4.40 & 49.60 & 9.43 & 0.987 & 50.23 & 54.69 \\ \hline 0.95 & 44.0 & 0.50 & 4.86 & 4.64 & 49.36 & 10.00 & 0.982 & 50.25 & 54.97 \\ \hline 1 & 44.0 & 0.00 & 5.11 & 4.89 & 49.11 & 10.58 & 0.977 & 50.26 & 55.26 \\ \hline \end{array}$

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