I haven’t made much progress on the calculations but I now think that the spider is not a Western Parson Spider after all but that it is a juvenile (Hogna antelucana), a species of Wolf Spider.
Better you should get us a picture.
I already set it free.
This post is too verbose to formulate a coherent answer. Ethylene glycol (EG) / water thermodynamic properties are frequently used by Chemical Engineers and a commercial simulator is used to offer some answers.
The density method is COASTALD (cite ) and the thermodynamic method used is CPA Equation of State (cite ). These are the popular methods in the industry however they can have inaccuracies.
If you take 70 gallons of pure water and 30 gallons of pure ethylene glycol at 68 F and mix it adiabatically, you will get :
99.6 gallons of mixture at 79.6 F. The EG volume fraction is 0.2980.
When this mixture is cooled to 68 F, you get 99.12 gallons of mixture. the EG volume fraction is same as before (obviously).
To the above cooled mixture, you can add 40.68 gallons of pure EG at 68 F, (adiabatically) to get a mixture with EG Volume fraction of 0.5 and the mixture temperature will be 73.68F. The volume of this mixture is 139.4 gallons. When cooled to 68F, the volume of the mixture is 139 gallons.
Since you have the ratios, you can easily work out how to fill a 16 liter volume.
Enjoy.
Why is that obvious? It’d change if the volume of the mixture changes, and from what you say in the next paragraph, the volume does change when the temperature changes.
Agreed Chronos - It does not make sense but hey it is tradition :smack:
The Liquid volume fractions are calculated at standard conditions (US system) which are different from Normal conditions. (SI system) 
As first stated by Frankenstein Monster in a previous post, the answer that the textbook author expected was approximately 4.57 liters. I finally found the specific gravities of 30 % by volume, 50 % by volume, and 100% by volume mixtures of ethylene glycol and water at 68 deg. F. By converting the specific gravities to densities I could then calculate the concentrations by weight from the concentrations by volume. After that, the calculations were pretty straightforward (shout out to excavating (for a mind). Converting the answer from a weight back to a volume I came up with approximately 4.73 liters. This differs from approximately 4.57 liters by approximately 0.16 liters. Using units more likely to be recognized by the SDMB, 0.16 liters is about 5.4 shots or about 3.6 jiggers.
Cite please ?
http://www.ppe.com/pdf/SpecificGravity.pdf
I had to interpolate to get 30% and and 50 %. Also, these are at 70 degrees F. and not 68 degrees F. but they are close enough for this exercise since the final answer is only being reported to three significant figures. And we could change the initial assumption of 68 degrees F. to 70 degrees F. if you are bothered by that.
All of the really comprehensive and reliable sources cost money, even online, and I am too cheap for that.
The title of your thread was “Negative Excess Molar Volume” (or, as you say back in Arkansas, f***ed up) .
And yet in your final calcs you ignore that totally. What’s the point ?
That’s what we did show. Due to the negative excess molar volume, we had to remove and replace 0.16 liters more volume than we would have otherwise expected.
Or to put it more simply, the difference is sixteen times greater than the implied precision of the problem, and so at that level of precision, ignoring the negative excess molar volume is not a justifiable approximation.
For fun, let’s redo the problem, but assume we are pouring in the pure antifreeze at the same time as the antifreeze/water mix is being drained. We can keep it simple and assume the antifreeze/water mix is always perfectly mixed.
Ready, go…
I don’t see where the implied precision of the problem is identified other than +/- 1 liter (the original problem does not say the automobile radiator contains 16.00 liters, but 16 liters). So, based on the implied precision of the problem (as well as real life), a difference of less than 6 ounces in 4 gallons is pretty insignificant. It certainly isn’t going to be the difference between your coolant freezing on a cold winter night, or not.
And, while we are nit-picking the problem, it must be a rather large and unusual automobile that has a radiator that will hold 16 liters. That’s one big honkin’ radiator! Heck, most modern automobiles (modern in this sense meaning not antique) typically have a overall cooling system capacity between 8 and 12 liters. It is true that a modern pickup truck with a V8 engine and towing package will probably have a cooling capacity of about 16 liters or more, but even then, the radiator is probably only going to hold half that, or less.
Now, back when cars were bigger, and nearly every US-built car was available with a V8 engine, a coolant capacity of 4 gallons was often considered typical. I remember Prestone commercials in the 1970s urging you to pick up two gallons to provide a 50/50 mix “for most automobiles”. This suggests to me (and, I could be wrong) that this is a recycled problem. That is, the original problem was probably a capacity of 4 gallons and to make it more “relevant” it was converted to liters. The person “modernizing” the problem probably just figured 4 liters per gallon, which is where the 16 liters comes from. It does not surprise me that the author of the problem didn’t understand the difference between the capacity of the radiator and the capacity of the cooling system.
Every measured value has an uncertainty associated with it and ideally all expressions of measured value should include the associated uncertainty. However, as is often the case, the author of the textbook did not include any uncertainties in what must be the measured values of 16 liters, 30 %, and 50%… The usual way to address such omissions is to assume that the last digit in the expression is accurate to within + or – 1. Thus we assume when the author of the textbook wrote “16 L” that this meant a quantity somewhere between 15 liters and 17 liters. “16 L” is said to contain two significant digits. The expressions “30 %” and “50 %” are ambiguous. The zeros may or may not be a significant digits. If the zeros were not significant then “30 %” means a value between 20 % and 40 % and “50 %” means a value between 40 % and 60 %. I find this absurd, so for the rest of my discussion in this post I will assume the zeros are significant and “30 %” means a value between 29 % and 31 % and “50 %” means a value between 49 % and 51 %. All of the other measured quantities that I found online concerning specific gravities and densities had at least five significant digits. Just so you know, I am aware that there are much better ways of assessing the propagation of uncertainties in calculated values than what is sometimes called the significant figure rules. But the significant figure rules are probably the only thing that a freshman non-science major will ever be exposed to and they are simple and convenient. Those are the rules that I will use for the remainder of this discussion. Using the significant figure rules, the answer of the problem as the author intended would be 4.6 liters and the answer to the problem accounting for the negative excess molar volume would be 4.7 liters. Thus, the difference between the two answers is “significant” in a technical sense.
Does this matter to an amateur mechanic working on his old truck in the backyard? No. 4.6 liters would probably be interpreted as “about on old milk jug full plus whatever I spill.” In my defense, I addressed this issue in my original post when I stated, “There will certainly be some critics who say that it is silly to worry about negative molar volume when adjusting the antifreeze concentration in your personal automobile. They would be correct. Nonetheless, I am pretty sure that the answer to the problem will include a negative excess molar volume that is easily measurable by readily available measuring devices (but maybe not readily available at your house). What if this were part of some critical and expensive research project concerning automotive cooling systems and precise and accurate information was essential? What if we were talking about 10 million radiators?”
On a side note, the uncertainty in a measurement is sometimes called the error of the measurement. In this sense, error does not mean mistake. Years ago I learned to never use the word error when discussing measurements with production management. Inevitably they will ask, “So it’s wrong then?” Also, avoid the word uncertainty because they will ask, “So you’re not even sure what you are doing?” I worked with a woman who said she had some luck describing the concept as the diversity of the measurement. Production management was afraid to question it because they were afraid they might appear politically incorrect.
I was basing that on Ynnad’s calculations to four significant figures, not noticing that the original didn’t have all those digits.
In every instructional lab ever, there’s the question “What were some possible sources of error in this experiment?”. I have to make it very clear to my students that “we screwed up” is not an error; it’s a mistake. Error is inevitable, but mistakes can and should be avoided.
Chronos said: “I was basing that on Ynnad’s calculations to four significant figures … .”
I don’t think I ever published a calculated value with more than three significant figures. But as we discovered, it can be argued that three is still too many.