## Learn about Junior Chemistry: Solutions 4 Video

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Learn about Junior Chemistry, Solutions 4, in this comprehensive video by bannanaiscool.

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Male Speaker: In the previous section, a lot of the things that we talked about were qualitative ways of describing a solution. But how about the quantitative ways? Well, here comes the math again. So, we need to able to describe how concentrated or dilute a solution is with numbers, and so we've chosen to describe concentration most conveniently with the moles of the solute dividing it by the liters of the solutions and you'll get moles per liter liters and that's concentration. So, concentration is moles of solute divided by liters of the solution. Here is a question that we can work on with and we can flush out here. If you have 10.0 grams of sodium chloride and 200 milliliters of solution, what's your solution concentration? We don't have moles right now, but we can find them. And then, we do have - well, we've got volumes, but it's in milliliters, we need to put in liters. Let's do all that in one step. You might be given formulas in your course in order to do these calculations. I guess they are okay, but you'll really need to memorize the formula as long as you know that its moles over liters that you need. So look at this. If you've got 10.0 grams of NaCl, well how do you convert that to moles? Well, the molar mass of NaCl is 55.65 grams for everyone mole. When you multiply those two numbers together, which is really 10 divided by 55.85 grams canceled left with moles of NaCl, you got moles in your numerator. You want liters in your denominator - well, I put the liters in the denominator. Now, look at this see everything is multiplied in the one line, but you really want to divide it by liters and you just put it in the denominator here; in the numerator just put the number one. So, 200 milliliters there, but now as moles per milliliter, you know what? Milliliters. You want milliliters to cancel on and leave yourself with liters and isn't there 1000 milliliters in one liter? Sure there are. So, look at that one line straight across to be able to do it and all you have to do in your calculator is punching 10 times of 1000 divided by 55.85, divided by again punch that in, divided by 200. And you are going to get 0.895 moles per liter of NaCl and that's the concentration. How you find the concentration when you are given the mass and the volume of the solution. You are going to be asked to prepare a solution and you are going to say well, okay that's easy enough. I'll just get the stuff and I'll put it together and teachers going to say oh hey, before you do that here is the concentration I want, here is the volume. You got to find the mass of the chemical that you need to weigh out. So, just remember you can take all this information for this question, 300 milliliters you want of 0.455 moles per liter potassium iodide. What's the mass of the potassium iodide that you need to make the solution? You've got volume and moles per liter. So volumes you can use that volume to cancel out liters here leave yourself with moles of KI. Once you have the moles of potassium iodide, you should be able to find the mass. So, look how it is done? If you have 300 milliliters of that solution, -- well, you know milliliters is great except that you can't use it in the calculation, so we convert the milliliters to liters. Now, some of you guys say hey look, I'm smart enough to just write 0.300 liters because I know that, that's what 300 mls is. Good for you. You go ahead and do that, but don't use your significant digits, you still need three. So, there is your milliliters canceled and leave yourself with liters. Look you are lucky, and then I've got the concentration in moles per liter. When I multiply moles per liter after the milliliters cancel, now the liters cancel and I have done moles of KI, don't want moles of KI, we want massive KI. And so, what you got to do is multiply it by the molar mass of potassium iodide, which of course is a gram per mole. There are the moles cancel and you are going to be left with grams of KI. And when

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