ok today we're gonna try some dilution problems. so the first one says 0.850 liters of a five point zero zero molar solution of sodium chloride is diluted to a volume of one point eight zero liters with water what is the concentration of the diluted solution? the next one asks what volume of 0.12 molar HCl can be prepared from eleven milliliters of a stock solution of 0.45 molar HCl? and the third one asks what volume of 1.59 molar NaOH is required to prepare a five point zero zero liter solution of 0.1 molar NaOH? so if this seems confusing go ahead and check out my tutorial on solutions and dilution it goes through these calculations it'll give you the information you need and when you're ready give this a try. so let's take a look at this first question now for all of these questions any time we're doing dilution we're always going to use this equation m1 v1 equals m2 v2 that means the concentration or molarity of the initial solution times the volume of the initial solution will be equal to the concentration or molarity of the final solution times the volume of the final solution so this is very easy to do we're just going to plug in what we know and solve for what we don't so in this particular case what is it that we're trying to get well we are asked what is the concentration of the diluted solution so we had some solution some initial solution and then we diluted it to get a different solution and we want to know the concentration of that diluted solution so we want em to that's what we're going for so let's solve for m2 just to make this easy on ourselves if we divide both sides by v2 we're going to get two equals m1 v1 over v2. now we just plug in what we know so m1 or the concentration of the initial solution is five molar so that's our stock solution of sodium chloride we had five molar sodium chloride that's m1 v1 is zero point eight five zero liters that's the initial volume that's the volume of that stock solution we used then we diluted it so when we diluted it we diluted it to v2 we diluted it to one point eight zero liters with water that's the final volume or v2 so then we just put these in a calculator and what do we get two point three six molar so this was diluted which means our answer must be of a lower concentration than the initial concentration so if you want to double-check and see that it makes sense we had a more concentrated solution we ended up with a diluted solution it should be a smaller concentration and here that is the case so this checks out that makes sense now moving on to the second one once again we are going to use m1 v1 equals m2 v2 so what is it that we're trying to solve for well it's asking what volume of a particular concentration solution of HCl can be prepared from a stock solution so we are given information about a stock solution and we want to know what volume of a diluted solution we can prepare so that's v2 we want to solve for v2 so let's just divide both sides by m2 and we're gonna get v2 equals m1 v1 / m2 and now once again we just plug in what we know so v2 is going to be equal to m1 so it says that the stock solution was 0.45 molar so that is our initial concentration and v1 is eleven milliliters because we used eleven milliliters of that stock solution so that's M 1 and V 1 and then we divide by M 2 that's the concentration of the desired diluted solution so we're trying to get a 0.12 molar solution we started with 0.45 and our final diluted solution needs to be zero point 1/2 molar and so what volume is our diluted solution going to be we plug those numbers in and we get 41 milliliters and the diluted solution needs to have a volume that is greater than the stock solution right because we're adding water to it so we want to check and make sure that's larger than the stock solution so we had 11 milliliters now we have 41 milliliters that is a greater volume so that checks out and that's going to be v2 for this problem now let's take a look at number three once again we're gonna use m1 v1 equals m2 v2 so what are we solving for this time well it asks what volume of a particular stock solution is required to prepare a diluted solution with those values so we're solving for v1 this time we want to know what volume of the stock solution we need to dilute to get to what we want so let's solve for v1 we just divide both sides by m1 and we get v1 equals m2 v2 over m1 so let's plug some things in v1 is going to be equal to 0.100 molar that is the m2 that is the concentration of the desired diluted solution so we're trying to make a solution that is 0.1 molar and so that's m2 v2 is the volume associated with that diluted solution so that solution that we're trying to get that's 0.1 molar we need five liters of it we want five liters of 0.1 molar solution so 0.1 times five that's our m2 and v2 and then dividing by M one that is going to be one point five nine molar that is the concentration of the stock solution so now we're trying to find v1 which is the volume of the stock solution we need and if we do the arithmetic we are gonna get zero point three one four liters so we are diluting the stock solution quite a bit so we want to make sure that the volume of the stock solution is much much smaller than the volume of the resulting diluted solution right that diluted solution is five liters and we're taking just a little bit of the stock solution 0.314 liters of it and diluting it up to five liters with water so that checks out and that's the answer for this question.