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Molarity and Dilution G23

Posted: Sat Oct 05, 2019 6:21 pm
by Rodrigo2J
G23) A lab technician has made up 100.0 mL of a solution containing 0.50g NaCl and 0.30g KCl, as well as glucose and other sugars. What is the concentration of chloride ions in the solution?
I understand that I need to find a solution that is in L/mol (molarity) and to do that, I have to convert 0.50g NaCl and 0.30g KCl into moles of Cl-. I'm just wondering if we assume that only these two molecules are the ones we are using to calculate the molarity or if we have to use glucose for some kind of conversion factor to solve.

Re: Molarity and Dilution G23

Posted: Sat Oct 05, 2019 7:05 pm
by Katherine Wu 1H
just assume that NaCl and KCl are the only two you're using

Re: Molarity and Dilution G23

Posted: Wed Oct 09, 2019 12:44 am
by Sofia Barker 2C
The question is asking for the concentration of chloride ions in the solution. The solution is made up of NaCl, KCl, and sugar. Sugars do not contain any chloride ions, thus it is implied that the chloride ions within NaCl and KCl are the only ions to take into account when solving for concentration.

Re: Molarity and Dilution G23

Posted: Wed Oct 09, 2019 9:36 am
by Chem_Mod
You may assume that the sugars do not contain chloride ions, they also do not give you amounts of the sugars so you cannot find moles of each either. One thing to note though is that molarity is mol/L, not L/mol.