Solutions: Mass Percent Example 2

by Jules Bruno
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determine the polarity of a sulfuric acid solution that is 5% sulfuric acid with a density of 50.9918 g per mole leader. Alright, so remember, Polarity itself is capital m. So we need to isolate the units of moles over leaders in order to isolate polarity. Alright. So remember that when we are given a mass%, 5 sulfuric acid here really means 5g of sulfuric acid Per 100 g of solution. And here we have the density of the solution, which is .9918 g of solution per one mL of solution. All right, so how do we use this information to help us get to our polarity? All right, so we're gonna start out with our mass%, 5.0 g of sulfuric acid Per 100 g of solution. What? All right. And we're gonna say here we want to get rid of the grams of sulfuric acid and isolate moles of sulfuric acid. So for every one mole of sulfuric acid, The molar mass of it is 98. g grams. You're cancel out now. I have Mosul sulfuric acid over grams of solution. Next, I need to get rid of grams of solution. Because remember, Polarity has leaders on the bottom. We do this by using our density. So here are density is .9918g of solution Per one million of solution. So here grams of solution, cancel out. And now I have milliliters of solution. I'm almost there. All I gotta do now is just change these milliliters into leaders. So remember that one Milli is equal to 10 to the negative three leaders. And like that, we've isolated moles over liters. So we've just isolated are more clarity. So we plug that in, we're gonna get .5056 Mohler. Now, since this has two sig figs will just bring this down into six things as well. So this is 60.51 Moeller. So that'll be our polarity for this given solution of sulfuric acid