# سوالات مخلوط

سرفصل: بخش ریاضی / سرفصل: مسائل کلمه / درس 12

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### دانلود اپلیکیشن «زوم»

این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید ## Mixture Questions

Mixture problems. Some word problems concern mixing solutions of various concentrations. What exactly does that mean? I will give an extremely concise introduction to some chemistry that may help you make sense of this topic. First of all it’s a fact that many different materials in nature happen to dissolve in water.

This forms a solution something dissolves, and it’s a fact that chemists often dissolve these materials that can be salt, sugars, acid and bases in water. The dissolved substance is called the solute. The concentration indicates how strong the solution is, that is, how much solute is dissolved in the given quantity of water. On the test, concentration will always be expressed as a percent.

So concentration equals the total amount of solute, divided by the total amount of solution and this is times 100%. Notice that concentration is not the ratio of solute to water but the ratio of solute to the total amount of solution, that is the amount of water plus solute. So it’s not the ratio of the two materials involved, it’s solute to the whole, that’s what constitutes concentration.

Some other details, just things you don’t have to worry about, but just so you know you don’t have to worry about these, sometimes concentrations on the test are given by as a percent by weight and sometimes by volume. This will be consistent so you don’t need to pay any attention to that, you can ignore that. Sometimes rather than water scientists use some other liquid, alcohol or something else the, that material that happens to be called the solvent it doesn’t matter what the solvent is, it doesn’t change anything about our calculation.

So all those are things that if you notice them you don’t need to worry about them they don’t matter. Okay, here’s a very easy question, pause the video and then we’ll talk about this All right. So HCl and again all these chemicals that are mentioned you have, you don’t have to have any idea what they are, as it happens HCl’s hydrochloric acid.

How much HCl and how much water must we use to create 5 liters of a 30% HCI solution? So this is a very basic question. So, the amount of HCl would be 30% as a decimal 0.3 times the whole, which is 5 liters and when we multiply out we get 1.5. So in other words we’re gonna need 1.5 liters of pure HCl and that means that the, the rest of it which would be 3.5 has to be water, so 1.5 is HCl, 3.5 is water, we combine those, we get a 30% HCl solution.

Now this question by itself is far too elementary to be a test question, but you will need these skills in more complex problems. This is much like we saw in distance, rate and time problems just given a simple here’s a distance, here’s a rate, find a time, well that’s too easy. That’s not gonna be a question by itself, but that is a skill you would need in solving a larger question and much in the same way you will need these skills in solving a larger question.

Here’s another practice question, pause the video and then we’ll talk about this. Suppose we start with 5 liters of a 30% HCl solution. How much water must we add to create a 20% solution? All right. Well, this is interesting. So the first thing we’re gonna figure out is how much solute do we have?

Again, 30% which is 0.3 times 5 we have 1.5 liters and of course, that’s not gonna change if all we’re adding is water. So we’re gonna have a new amount, and it’s gonna be 20% and it’s gonna have the same amount of HCl in it. The same amount of solute in it and so here we need to solve for X and that would be, be the total amount of solution.

So of course 0.2 is 1.5 multiply both sides by 5 we get 7.5 liters, and so the old solution was 5 liters. The new solution is 7.5 liters, so the difference is 2.5 liters and that’s how much extra water we need to add, to bring this from a 30% solution, to a 20% solution. In that problem, we added water to the solution, to make a less concentrated solution.

Notice we could also add pure solute to a solution to make a more concentrated solution. The test could ask about either of those scenarios, but the most common questions involve mixing two solutions of two different concentrations. So here’s a practice question along these lines, pause the video, and then we’ll talk about this.

Okay, again you do not need to know what that chemical is, HCPO4 that happens to be phosphoric acid, it’s an ingredient in Coca-Cola, but at any rate, suppose we start with 8 liters of a 60% solution. We add 4 liters of a C percent solution. So C is the unknown and the result to 12 liters of a 50% solution. What is C? Well, the first thing I’m gonna do is figure out how much solute did I begin with and how much solute did I end up with?

So in the first solution, I had 4.8 liters of solute. In the result in solution, the final solution, I add 6 liters of solute and that means that the amount of solute that was added, that amount of solute was 1.2 liters. So let’s think about this, there’s 1.2 liters of solute in that 4 liter solution. So, we just have to fi, figure out 1.2 is what percent of 4?

So divide this out, and we get that this is 30% and so C equals 30%. Sometimes in solution problems, if amounts of two different solutions are initially unknown, we have to set up simultaneous equations. This is probably the most popular solution problem type that you’ll see on the test.

One equation will always be a total equation that is the total volume or the total mass or weight. The other equation will always be about the amount of solute. So those are the two equations you’re always gonna set up and once you have these two equations, you use the techniques for simultaneous equations. Here’s a practice question, pause the video and then we’ll talk about this.

Okay, here we have a question about H2S04, sulfuric acid, you don’t need to know about that. Suppose we start with an, with unlimited supplies of a 20% solution and a 50% solution. We combine X liters of the first, with Y liters of the second, to produce 7 liters of a 40% solution, what does X equal?

So clearly, X plus Y equals 7 that’s one equation that we have, we have X liters of the first thing, Y liters of the second thing, so X plus Y equals 7 that’s one equation. So what we’re gonna do is focus on the solute, the X plus Y equals 7, that’s the total equation. The solute, well, the resultant solution is 40% of 7 liters so that’s 2.8 liters of solute.

All right, so that our total equation is X plus Y equals 7. The solute from one is 0.2 times X. The solute from two is 0.5 times Y. We add those two, and it should add up to the total solute. So that’s our second equation. Our first equation is X plus Y equals 7.

Our second equation is this solute equation, 0.2X plus 0.5Y equals 2.8. So in that first equation what I’m gonna do, I’m just gonna solve it for Y. The reason I’m gonna solve it for Y is that when I substitute, that will eliminate Y and it will leave me with an equation for X, which is what I want. So I’m gonna plug this into the equation, into the second equation, substitute it in.

I’m gonna multiply everything by ten, just so I don’t have to deal with decimals. I’m gonna distribute, collect the terms, simplify a bit, subtract 35 from both sides we get negative 3X equals negative 7 divide by negative 3 we get X equals seven-thirds liters, and that’s the answer to the question. Concentration is the ratio of solute to the total solution most often expressed as a percent, and here of course we have to be very fluent in going back and forth between percents and decimals.

We can change concentration by adding water or adding pure solute to a solution, or by mixing two solutions of two different concentrations. If we mix two solutions in unknown amounts to get a total known, a total, a known total of a known concentration, we set up simultaneous equations. One equation is the total amount equation and the other equation is for the amount of solute.

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