Homework Problem G.23

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Marni Kahn 1A
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Homework Problem G.23

Postby Marni Kahn 1A » Mon Sep 30, 2019 3:27 pm

A lab technician has made up 100 mL of a solution containing .5 g of NaCl and .3 g of KCl, as well as glucose and other sugars. What is the concentration of chloride ions in this solution?

I understand that you must divide by the amount of moles of Cl by .1 L, but I do not know how to determine the amount of Cl in this solution. I converted .5 g of NaCl and .3 g of KCl to moles, but I am unsure of how to calculate the amount of Cl moles alone.

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Re: Homework Problem G.23

Postby Chem_Mod » Mon Sep 30, 2019 3:33 pm

Use the chemical formulas given in the problem, KCl and NaCl. This tells you that in each ionic compound of KCl, there is a 1:1 ratio of K+ to Cl-, and same goes for NaCl. Therefore, dissolving one mole of KCl will yield one mole of Cl-, and same for NaCl. You would thus find the concentration of Cl- by combining the amount of Cl- resulting from dissolution of each compound.

Brooke Yasuda 2J
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Re: Homework Problem G.23

Postby Brooke Yasuda 2J » Tue Oct 01, 2019 12:06 am

Yes, so as the previous post says because the solutes NaCl and KCl only have one Chlorine per molecule, a mole of NaCl or KCl will also give you a mole of Chlorine. Therefore, just convert the grams of NaCl and KCl to moles and from here, because the solution contains both the NaCL and KCl in a 100 mL flask, you just have to add the moles of Chlorine that you calculate and then divide by total volume to find the molarity.

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Re: Homework Problem G.23

Postby MMckinney_4H » Tue Oct 01, 2019 12:22 pm

If you use the periodic table to find the molar masses of each of the elements, you find that Na has a molar mass of 22.99, Cl has a molar mass of 35.45, and K has a molar mass of 39.098. Adding Na's molar mass with that of Cl, you get that 1 mol of NaCl has a mass of 58.44 g. To find how much NaCl we have in mols in .5 g of NaCl, we can use our ratio of 58.44 g / 1 mol = .5 g/ x mols. We find that we have .0085558 mols of NaCl in .5 grams. Because Na and Cl have a 1:1 ratio in NaCl, for every mol of NaCl, we have one mol of Cl. Similarly, we have .0085558 mols of Cl in .0085558 mols of NaCl. If we do the same for KCl, we find that one mol of KCl has a mass of 74.548 grams. So, .3 grams of KCl has .004024 mols of KCl and likewise, .004024 mols of Cl. If we add the amount of moles of Cl from each, we get that .004024+.0085558= .0125798 mols of Cl in the solution. To find the molarity, we simply use the formula Molarity = Moles of Solute/ Liters of Solution. We are given that there are 100 mL of solution. To use the formula for molarity, we convert this to .1 liters of solution altogether. Then just plug everything in to find that the molarity is .1258 or .13 M Cl ions.

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