the density of 63.7 weight percent sodium hydroxide is 0.915 g per milliliter, how many milliliters of water should be diluted to 850 mls to create 8500.4 to 5 molar sodium hydroxide. Alright, so here they're asking us how many ml of water? It's really asking how many milliliters of solution should be diluted. So that's what we're looking for. What we're gonna do now is we're going to isolate all the information and see how we can isolate those ml of solution. We're told here that we have 63.7% sodium hydroxide. So that means we have weight% 63.7 g of sodium hydroxide for every 100 g of solution, We're told that the density of the solution is .915. So that's .915g of solution per one millimeter of solution. What else are we given? We're told that we have 850 mls to create 8500.4 to 5 molar sodium hydroxide. So we have 850 mls And then remember polarity is moles over L. So that's just .4-5 moles of sodium hydroxide per one liter. We're gonna start out with 850 mls first because remember that's just one unit easier to deal with. The others have a mixture of two units involved, harder to um to control. So we have 850 mls, This 850 MS is attached to this polarity. If we could change those mls into leaders and multiplied by the polarity, we can isolate the moles of sodium hydroxide. So we're gonna change mls into leaders. Remember that one leader is 1000 ml. Now that we have leaders of solution, we can cancel out the leaders from the polarity given. And that'll give us moles of N A. O. H. At the end. But here we gotta keep going. So so far we've used this portion of the given information. What do we have left? We have left the mass or weight percent and we're left with density. We know that we need to isolate ml of solution which are right here. Next I see that I have moles of N. A. O. H. So if I could change them into grams, I can cancel out these grams of N A. O. H. So we're gonna stay here for every one mole of N A. O. H. What is the mass of any O. H. Well, it's made up of one sodium, one oxygen and one hydrogen. And based on the atomic masses from the periodic table. Each of them is 22.98977 g. 15.9994 g and 1.00794 g. Add them all up together gives us 39.9971 g. So we have grams of N A. O. H. So bring down the weight percent, 63.7 g of any O. H. R. On the bottom, 100 g of solution on top. Finally I can cancel out the grams of solution by using the density of the solution given. So we put .915g of solution here on the bottom And then one middle liter of solution here on top. So when we punch all that in that gives me 24.79 ml of solution. And here we have three sig figs, three sig figs, four sig figs and three sig figs. So we can run that just 2 24.8 ml of solution would need to be diluted 2 850 mls in order to create 8500.4 to 5 Mueller and A O. H solution. Again, the warning of some of these problems can be a bit challenging. But the best approach, like I've been saying is to write down what you're looking for, first write down all the given information, try to isolate the units that you need to get the desired answer at the end. Doing that helps us to navigate through all the wording, all the crazy jumbled numbers in order to get our final answer. So hopefully guys were able to follow along in terms of these typical types of dilution questions and we'll continue onward into talking more about chemical reactions as well as Tokyo metric calculations that will become a focal point in discussing concentrations of solutions