Skip to main content
Ch.13 - Solutions & Their Properties

Chapter 13, Problem 78a

A 0.944 M solution of glucose, C6H12O6, in water has a density of 1.0624 g/mL at 20 °C. What is the concentration of this solution in the following units?(a) Mole fraction

Verified Solution
Video duration:
4m
This video solution was recommended by our tutors as helpful for the problem above.
1165
views
Was this helpful?

Video transcript

Hello. Everyone in this video, we're being asked what is the mole fraction of sucrose in a solution? If the solution has a density of 0.997 g per milliliters. So go ahead and recall what maturity is similarity is equal to our moles of solute over the leaders of solution. Let's go ahead and assume this problem that we have one liter of solution. Alright, So for our first calculation we can calculate for the mass of solution. So we said that we're assuming that we have one leader solution. So we're gonna go ahead and serve the dimensional analysis with that. We want to convert our leaders into milliliters. So for every one leader we have 1000 mL and then we go ahead and use the density that's given to us. So that's 0.997 g for every one millimeter. So you see now that the leader will cancel as well as the military leader. And once you put all these values into the calculator, we get the value to be 997 g. Now next, we're gonna go ahead and calculate for the mass of our salute. So here we're starting off with the maturity that's being used and that's 0.819 moles. And we can just assume that since over one liters anyways and then we can go ahead and use the molar mass to give us grams. So the molar mass. Is that for every one mole of super gross. We have 342.3 g. Well, should make this a little bit lower. So we cancel everything out. All right, So if we do so we see the units of mole will cancel. And for our grams, the value for that is to 80.3437 g. Lassie cackling for the mass of water. So H 20. That's just 997 g of the mass of the solution minus the mass of our salute. So to 80.3437 g. We put that. We got a difference of 716. g scrolling down here for more space. So for our moles of sucrose, we established that to be 0.819 moles. Now for the molds of our water, H 20. Starting off the mass of that 716.6563 g. Using the molar mass of water, which is 18.2 g per mole. You see here that the g will cancel, leaving us with the units of moles, which is what we want. So we get the product here of 39.77 moles of water. Now, finally, we can use this information in purple that we just software to calculate for our mole fraction of sucrose. So here we want on the numerator, numerator to be the moles of are so gross. So that 0.819 moles. And then our denominator will be will be the total moles of our solution. So that's so gross and the H20. So that's just adding 0.819 and 39.77. So once you get that you get the value of 0.020177. So if you run that to the correct number of significant figures will get the final mole fraction of sucrose to be 0.0202. So this right here is going to be my final answer for this problem.