If cost per gram were not a concern, which of the following subst...

Question

If cost per gram were not a concern, which of the following substances would be the most efficient per unit mass for melting snow from sidewalks and roads: glucose 1C6H12O62, LiCl, NaCl, or CaCl2? Explain.

Accepted Answer

Hello. Everyone in this video we're being asked if magnesium chloride, urea, potassium chloride or sodium acetate will be the most efficient and multi ice from parking lots and we're disregarding economical concerns here for the sake of this problem. So we're basically being asked to calculate the numbers of moles of solute particles and we can go ahead and assume that we have one g of each substance starting off with your area here. This has a molecular formula of CH four and 20. Which has a molar mass of 60. g per mole. So the moles of soil particles is equal to. Let's see. So we have one g of this and we're gonna go ahead and multiply this by the molar mass to give us the moles. So for every one mole of Yuria we have 60. g. We can see here that the gram units will cancel. So this gives us an answer to be 0. malls of saw you particles. So iria does not dissociate into ions because it is a non ionic substance. Let's go ahead and highlight this as well. So you can say that it will not dissociate because it's a non ionic substance. So that's something to keep in mind. Next we go ahead and do potassium chloride. So that's K. C. L. This has a molar mass or a molecular formula let's see you. So potassium chloride. So we have that to be K. c. l. and smaller masses 74.55 g per mole. So for the moles of solute particles again we have one g of this. I'm gonna go ahead and multiply this by the molar mass to give us moles. So for every one mole of K C. L, we get 74.55 g. So once you put this into a calculator we see that we get a value for potassium chlorate to be 0.0268-8 moles. Alright, so protesting chloride associated to to solute ions per formula unit in the solution has shown in the dissociation equation which is just K. C. L. Dissociating into K. Plus and cl minus. So thus there's going to be that the number of moles of solute particles over test in florida is going to be definitely multiplied by two. Since we have two of these different ions. Let's go ahead and scroll down for more space earlier. We did yuri already and we did potassium chloride. Alright, so moving forward we now do sodium acetate So far. No sodium acetate. This has a molecular formula to be CH three C. O N. A. This has a molar mass to be 82. g per mole. Alright, so again, the more solid particles here We have one g of this and then we're gonna go ahead and multiply this by the molar mass. So for every one mole of sodium acetate we have 82.0 g. Of course the units of grams will cancel leaving us with the moles to be 0.024381 moles forgot to mention that we're gonna go ahead and multiply this by two ions because that's the ions that we will get the same thing as our K C. L. So again, sodium acetate associated into two solute ions per formula unit. And a quick solution. And this association formula is going to be that the CH three C 00. And A will dissociate into the sodium Canadians and the CH three C O minus an ion. Now, lastly we have our magnesium chloride. So that's MGCL two. And this has a molecular formula to be, let's see. So we have the molar mass of the MG cell to to be 95.21 g per mole just like before we go ahead and calculate for the molds of the solid particles. So given that we have one g of this, we're going to multiply it by the molar mass. So for every one mole of magnesium chloride we have 95. g. Again, the grams unit will cancel leaving us with the moles to be 0.0 scrolling down everywhere. Out this dissociation equation we get at the mg cl two will dissociate into MG plus carry ons. And for the sake of balancing out this kind of equation, we need two moles of our chlorine and ions. So because of this we have multiplied this here by three ions. So this answer has already been multiplied by R three ions. So again, magnesium chloride associated three different solute ions per from a unit in an aqueous solution as seen as this equation here. So the number of moles of solute particles for the magnesium chloride is going to be multiplied by three. So take a look at all these calculations here. Mg CL two produces the most solid particles program. Let's go ahead and get that in writing. So, MG CL two produces the most solute particles program Since we assume that we have one g of each already in our calculations and to be more exact these more values is 0.031509. And that's just the answer that we got over here and we're comparing this MG cL two for the solar particles amongst the mentioned substances. So therefore what we can tell from those answers that MG cL two is going to be the most efficient per unit mass and melting ice from parking lots. So what we just wrote here is going to be my final answer for this problem

If cost per gram were not a concern, which of the following substances would be the most efficient per unit mass for melting snow from sidewalks and roads: glucose 1C6H12O62, LiCl, NaCl, or CaCl2? Explain.

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