Homework Problem G23

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Alexia Joseph 2B
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Joined: Thu Jul 27, 2017 3:01 am

Homework Problem G23

Postby Alexia Joseph 2B » Thu Oct 05, 2017 7:16 pm

In the problem, "A lab technician has made up a 100.0mL solution containing 0.50g NaCl and 0.30g of KCl, as well glucose and other sugars.This question asks that we find the concentration of the chloride ions in the solution. How do you go about solving this problem/what steps do you need to take? How do you account for the Cl ions in both compounds?

Dylan Davisson 2B
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Re: Homework Problem G23

Postby Dylan Davisson 2B » Thu Oct 05, 2017 7:39 pm

To start, you want to find out how many moles of each molecule (NaCl, KCl) exist within the solution. So, for each, you divide the amount of grams present by the molecule's molar mass. Now you can multiply the moles of NaCl by the volume of the solution, then multiply the moles of KCl by the volume of the solution.
When you add those two products together, it leaves you with the molar concentration (molarity) of the chloride ions in the solution.

SantanaRodriguezDis1G
Posts: 56
Joined: Sat Jul 22, 2017 3:00 am

Re: Homework Problem G23

Postby SantanaRodriguezDis1G » Thu Oct 05, 2017 7:44 pm

You would start by using the 0.50g of NaCl given and divide it by the molar mass of NaCl (58.44g mol^-1) and then multiply that by the ratio of moles of NaCl to Cl ions which is 1:1.
And then you would add that to the concentration in KCl by starting with the given (0.30g KCl) and dividing that by the molar mass of KCL (74.55g.mol^-1 ) and then multiplying that to the molar ratio which is again 1:1 in this problem
That's how you would incorporate the concentration of Cl ions from both compounds.
So it would look like( (0.50gNaCl)/(58.44g.mol^-1) ) + ( (0.03g KCL)/(74.55g.mol^-1Kcl))= .013mol Cl
The next step would be to find the molarity and we know that M= n/c where n= amount in moles of solute and v= volume in liters
therefore you would divide the amount in moles of Cl ions by the given volume 100.0ml(which must be converted to (.1L)
M=(.013mol)/(.1L)
=.13 mol.L^-1 Cl
I hope this helps


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