A 0.944 M solution of glucose, C6H12O6, in water has a density of...

Question

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?(c) Molality

Accepted Answer

Hello. Everyone is on this problem being told that we have a 1.32 molar concentration of an adequate solution of ethanol. It has a density of 0.951 g per milliliter. So we're calculating for the morality of the solution. So let's recall the difference between what polarity is and morality. So familiarity. That's moles of our salute over liters of solution and then for morality as equal to the moles of our salute over the kilograms of our solvents. Alright, so in this problem we can go ahead and assume that we have one leader of solution. So from that we can go ahead and calculate for the mass of our solution. Starting off with our assumption here with one liter of solution, We convert this unit into ml. So for every 1000 ml we have one leader and then we can go ahead and use our density. That's given to us in the problem as a converted factors into grams. So for every one millimeter we have 0. g and then we can see that the units of leader and a leader will cancel. So the massive solution then is going to be g of our solution. Now calculating for the mass of our salute. So starting off with the polarity So that's 1.32 moles for every one liter of solution. We're gonna go ahead and cancel out the units of leaders with the assumption here of one leader and then continue our dimensional analysis. We want to convert to moles into grams to get the units of mass. How we do that is by using the molar mass of ethanol. And that's 46.7 g for every one mole of ethanol. And now we can see that the units of multiple cancel and leaders. So that's just a multiplication sign here. Right, this a little bit better. Alright, so putting all these values into vocabulary, I get the total of 60.81- g of ethanol. Now calculating for the mass of the solvent. That's going to be 951 g -60.81 - four g. That gives us a difference of 890. g of our solvent is H 20. So that's water. Alright. Now just scrolling down a little bit. So now we're calculating for the morality. So again, we have moles on top and that's just 1.32 moles. And we just got that from Our polarity. And then our denominator is let's see, we have 890.1876 g. We want to convert our g to kg. So we'll do a direct conversion here. So For every one kg we have 1000 g. And we see here that the grams will cancel. Should we need this in the bottom for it to cancel? So we have the grounds on the bottom. We do actually need it on top. So it can cancel this way. So now we see if we find the answer, we have the units of moles over kilograms. So putting this in the calculator will get the final value of 1. being our morality of the solution, and this is going to be my final answer for this problem.

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?(c) Molality

Watch next