A chemist needs to mix a solution that is 34% silver nitrate with...

Question

A chemist needs to mix a solution that is 34% silver nitrate with one that is 4% silver nitrate to obtain 100 milliliters of a mixture that is 7% silver nitrate. How many milliliters of each of the solutions must be used?

Accepted Answer

Hello today we're going to be solving a concentration problem using a system of equations. So the problem states that Alex is mixing a 20% concentration of orange juice with another orange juice, that has a 40% concentration. This mixture is going to produce a juice that has ml with a 32% concentration. We need to figure out how many ml of the 20% and 40% concentration. We need to produce this type of juice. So we are told that we're mixing two different types of juices. So we can let x represent our 20% juice container and we can let y represent our 40% juice container. So if we're thinking about this with respect to the just the concentration, We know that we have a container that has a 20% concentration and we're combining that with the container of 40% concentration And that's going to produce a third container that has a 32% concentration with a total of ml. So multiplying 32 with 250 is going to give us 8000. So we have 20 X plus 40 Y. Is equal to 8000. And that is gonna be our first equation. Now, if we think about this problem with respect to just the milliliters, we know that we have some unknown amount X. And we're combining that with an unknown amount Y. To produce 250 mL, we're gonna go ahead and let that be our second equation. So we can take these two equations and solve for X&Y. Using the method of systems of equations. So if we line this up, we get 20 x plus 40 y Equal to 8000 and we're going to add X plus y Equal to 250 to that equation. So what we want to do is go ahead and eliminate one of the variables to sulfur the other. Well, if we multiply our second equation by negative 20 what we end up with is 20 X plus 40 Y equal to 8000 plus negative 20 X minus 20 Y is equal to 250 times negative 20 which is negative 5000. And we're gonna go ahead and add these equations together while 20 X minus 20 X is going to cancel each other out and 40 y minus 20 Y. Gives us 20 Y 8,000 -5000 is going to give us 3000. And in order to solve for why we just need to divide both sides of this equation by 20. Doing so is going to give us why is equal 2 150 ml. So again, this is representing our 40% container. So this tells us that we want 150 ml of the 40% container. Okay, so we solved for one of the containers but now we need to solve for the other. How do we do that while what we can do is go ahead and take our value of y and plug it in to our second equation in order to solve for value X. Doing so is going to give me X plus because y is 100 and 50 equal to 250 In order to solve for X. All I have to do is subtract 150 to both sides of this equation, and X is going to equal to 100 million years and that is going to be our 20% container. So just to summarize, we are going to combine 100 mL of our 20% container with 100 50 mL of our 40% container to produce the 32% container. So our two solutions is X is equal to 100 Y is equal to 100 and 50. And the answer to this problem is going to be c. So I hope this video helps you in understanding how to approach this type of problem. And I'll go ahead and see you all in the next video

A chemist needs to mix a solution that is 34% silver nitrate with one that is 4% silver nitrate to obtain 100 milliliters of a mixture that is 7% silver nitrate. How many milliliters of each of the solutions must be used?

Key Concepts

Concentration

Systems of Equations

Mixture Problems

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