so here in the second half of section 6.3 molarity we're going to focus on being able to perform dilution calculations using the dilution equation so dilution is the process whereby the concentration of a solution is lessened by the addition of solvent and it's very common in chemistry to have stock solutions that are concentrated and then dilute them for use at a later time so as noted here dilution is a common means of preparing Solutions of a desired concentration and by adding solvent to a measured portion of a more concentrated stock solution we can achieve a particular concentration that we desire so um here's an example where we have two solutions they both contain the same mass of copper but the solution on the right is more dilute because it has a greater volume right the copper nitrogate nitrate is dissolved in more solvent so the molar amount of solute in a solution is equal to the product of the solution's molarity and its volume in milliliters so that's what's shown in this equation here number of moles equals molarity times volume in liters and we can see that that's the case by looking at the units so molarity has units of moles per liter and liters are and volume in liters obviously has the units of liters so that our leaders could cancel out and we're left with number of moles so expression like expressions like these may be written for a solution before which is uh represented by the number one and after represented by the number two a dilution occurs okay and furthermore since the dilution process doesn't change the amount of solute in the solution uh the number of moles at the beginning and one is equal to the number of moles of solute at the end okay so again just to be clear the number of moles that you begin with uh of solute is the same at the beginning of the process and at the end of the dilution process so what that means then is that N1 equals N2 and therefore we can set the two equations that we had earlier equal to each other so um we have that the molarity in the beginning times the volume in the beginning is equal to the molarity at the end times the volume in liters at the end and we can actually write a very general equation for any type of concentration units that we might want to use and for any volume units that we want to use so typically what we'll simply say is that C1 V1 equals C2 V2 meaning the initial concentration times the initial volume is equal to the final concentration times the final volume so let's see how we can use this to calculate a concentration or volume so here we have a problem it says if 0.850 liters of a five molar solution of copper nitrate is diluted to a volume of 1.8 liters by the addition of water what is the molarity of the diluted solution so what is the final molarity so the final molarity that we're interested in is C2 right so that's the thing we don't know um and so we what we can do is solve for that so dividing both sides by V2 we have that C2 equals C1 V1 divided by V2 and then we can put in our values the initial concentration is 5 molar so that goes here the initial volume is 0.850 liters so that goes here and the final volume is 1.80 liters so that's V2 that goes in the bottom here and then we simply work the math out notice that liters will cancel out and we'll be left with the concentration in molarity which turns out to be 2.36 molar now just a final word on this is we can use this equation to um to solve for any one of these four quantities right so as long as we know the other three then we can solve for the fourth so if you wanted to know for instance what volume you needed to achieve a certain final concentration we could solve for v2 or if you wanted to know um uh how much volume of a particular stock solution say C1 that you should use that would be V1 we can solve for V1 so this is a very useful equation