The density of acetonitrile (CH₃CN) is 0.786 g/mL and the density of methanol (CH₃OH) is 0.791 g/mL. A solution is made by dissolving 22.5 mL of CH₃OH in 98.7 mL of CH3CN. (c) Assuming that the volumes are additive, what is the molarity of CH₃OH in the solution?

Video transcript

Hey everyone, we're told that acetone has a density of 0.7845 g per mil leader and di ethyl ether has a density of 0.7134 g per minute leader a volume of 46.5 mL of acetone is dissolved in 1 20.5 millimeter of di ethyl ether, calculate the molar concentration of acetone in the solution. Assume the volumes are additive to determine the moles of our acetone. We can go ahead and start off with our 46.5 ml of acetone And we can use our density of 0.7845 g per mil leader for our acetone. And since we want to convert this into moles, we can go ahead and use acetone moller mass, which is 58. g per one mole of acetone. Now, when we calculate this out and cancel out all our units, We end up with a total of 0.6281 mole of acetone. Now let's go ahead and calculate our volume. Since our questions them told us that the volumes are additive. We can go ahead and start off with our 46.5 mL from our acetone and we can add our 1 20.5 mL from our di ethyl ether. This will get us to 167 ml And we want to convert this into leaders and we know that we have 10-3 ml per one leader. So this gives us a total volume of 0.167 L. Now to calculate the molar concentration of acetone, we can take our 0.6 to 81 mole of acetone And we can go ahead and divide that by our total volume of 0. L. This will get us to a molar concentration of 3.76 moller, and this is going to be our final answer. So I hope this made sense and let us know if you have any questions.