here. It's asking how many grams of 53.1 weight percent sodium chloride should be diluted to 2.50 liters to make 0.15 moller and a cl. Now in this question we're not dealing with M one V one equals M two V two because we only have one volume and one polarity um involved within the question what this question itself is asking? It's asking us for the grams of solution. So how many grams of solution are needed based on what they've told us. Now with this information, we're gonna put all the given information on the other side and organize it to help isolate our grams of solution. So we have 53.1 weight percent sodium chloride solution. So that means we have 53.1 g of sodium chloride Over 100 g of solution. There goes the grams of solution. We know we need to isolate at the very end of all our calculations, we're told that we have 2.50 liters and then we're given a polarity of 0.15. What does that really mean? While polarity is moles over liters. So 0.15 molar really means that I have 0.15 moles of sodium chloride Per one L of solution. So we know we need to isolate my grams of solution. We're gonna start out first with the 2.50 L because it's just one unit by itself, easier to manipulate. So we have 2.50 L. We want to cancel out leaders. So we're gonna bring down the polarity, we're gonna have the one liter of solution here on the bottom And we're gonna have .15 moles of sodium chloride here on top. Leaders cancel out. At the moment. We have moles of sodium chloride. We know that. We need to isolate these grams of solution to be able to do that. We need to cancel out these grams of sodium chloride. That means that the molds that I have right now have to change into grams. So the molecular weight that were given for sodium chloride we can use. Now We're gonna say one mole of sodium chloride on the bottom weighs 58.443 g of sodium chloride on top. So moles cancel out. Now I have grams of sodium chloride Now that I finally have those grams of NACL. I can use them to cancel out the 53.1 g of NACL. So we have 53.1 g of NACL. Here on the bottom. 100 g of solution on top. When we plug that in, that's gonna give me 41.27 33 g of solution. Now, if we look that's three significant figures. Three significant figures. Two significant figures. So at the end, we're going to say we need approximately 41g of solution to accomplish this. Now, taking a look at this example, let's see if you guys can attempt the practice problem that's left on the bottom, Come back, take a look and see how I approach that same exact question.