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The alcohol content of hard liquor is normally given in terms of the "proof," which is defined as twice the percentage by volume of ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)$ present. Calculate the number of grams of alcohol present in $1.00 \mathrm{L}$ of 75 -proof gin. The density of ethanol is $0.798 \mathrm{g} / \mathrm{mL}$.

$$3.00 \times 10^{2} \mathrm{g}$$

Chemistry 102

Chapter 12

Physical Properties of Solutions

Solutions

University of Central Florida

Rice University

Brown University

Lectures

03:58

In chemistry, a solution is a homogeneous mixture composed of two or more substances. The term "solution" is also used to refer to the resultant mixture. The solution is usually a fluid. The particles of a solute are dispersed or dissolved in the solvent. The resulting solution is also called the solvent. The solvent is the continuous phase.

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All right, So we have this liquid that is 75 proof. And according to the question, that means that we have Ah, 37 0.5% Canton off, uh, alcohol, ethanol by volume. So this is by volume. Okay, So the question is, how much of ethanol in mass we have in this letter of gin? So let's ride just for convenience. Convenience? That's right. 1000 middle letters. Okay, so we want to look at 1000 mill letters off gin. Okay? Gen. So the first thing is to apply this ratio here that would just we just talked about, So we want to take 37.5% off 1000. I'm gonna do this by writing this factor here. So I have 37 0.5 milliliters off ethanol. I'm gonna write e t o age. You know, I have that in 100 mill letters of June. Now, hopefully, that doesn't look strange. That's basically percentage. Okay, I really I re ruled that it is a fraction. So all I'm doing is the staking 37.5% of 1000 and that and that step will give you the volume off alcohol inside is 1000 milliliters of gin. So what's next? Well, we have the volume about Oh, by doing this calculation here, however, we want mass, and we were given the density of ethanol. So that's the last step. We have to apply the density off ethanol. So I'm going to write that as 0.7 98 grams off ethanol. Okay, O h, that is present in one militar off ethanol, right? This is the density. What I'm saying here is that for one middle litter of ethanol, I have this mass and this is the density. Okay, hopefully the look. That is clear now, Senator Chuck, let's check if the units make sense here. So volume of Jim, I'm canceling this out by using the percentage and then the volume of ethanol. I'm also canceling that out by using the density off ethanol. And at the end, we have the mass of ethanol, which is what we want. So we can go ahead and perform this computation, and you should get something like to 99 grams off ethanol

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