Module question

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705088777
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Joined: Fri Sep 28, 2018 12:18 am

Module question

Postby 705088777 » Wed Oct 03, 2018 12:18 am

A solution is prepared by dissolving 55.1 g of KCl in approximately 75 mL of water and then adding water to a final volume of 125 mL. What is the molarity of KCl(aq) in this solution?

I converted the KCL grams into moles. then I multiplied that by .075L and finally divided by .125L. I'm not getting the right answer though. What am I doing wrong

gillianozawa4I
Posts: 31
Joined: Fri Sep 28, 2018 12:27 am

Re: Module question

Postby gillianozawa4I » Wed Oct 03, 2018 12:29 am

The way I did the problem was to first convert 55.1 g of KCl to moles. The molar mass of KCl is 74.6 g/mol, so 55.1/74.55 is 0.739. Since the number of moles remain the same in a dilution problem, you need to just divide the number of moles by the number of liters, which is .125 L. 0.739/0.125 = 5.91 M.

Alma Carrera 3C
Posts: 33
Joined: Fri Apr 06, 2018 11:02 am

Re: Module question

Postby Alma Carrera 3C » Wed Oct 03, 2018 12:36 am

Yeah, so your mistake was multiplying the number of moles by .075 L. The 75 ml of water were used to dissolve the KCl, but the total volume was actually 125 ml. So all you need to do is follow the formula C=n/V, so you would divide the number of moles of KCl by the total volume in Liters, .125 L.

Ashley Kim
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Joined: Fri Sep 28, 2018 12:19 am
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Re: Module question

Postby Ashley Kim » Wed Oct 03, 2018 12:38 am

This might be repetitive, but I used the M1V1=M2V2 equation.

I converted the grams of KCl to moles and then found the molarity - then multiplied that value by the original volume (in liters). Then, I put in 0.125 L on the other side and solved for the second molarity. It's definitely quite repetitive, but it is easy to understand.

Let me know if you have questions!

Stevin1H
Posts: 89
Joined: Fri Sep 28, 2018 12:17 am

Re: Module question

Postby Stevin1H » Wed Oct 03, 2018 12:10 pm

For this problem, I approached it seeing the information that I was given in the problem. Given 55.1g KCl in approximately 75ml (0.075L) of water then adding water to a final volume of 125ml (0.125L), the formula M(initial)V(initial)=M(final)V(final) was needed to solve this problem.

First, I converted the sample of 55.1 KCl into moles by dividing by its molar mass. (Molar mass of KCl is 74.55g/mol). By doing 55.1g/74.55g.mol-1, I got 0.739 mol KCl. Given the moles, I used the formula to find the initial molarity.

Molarity (M)=moles of solute(n)/volume of solution(L)

Molarity (M) = (0.734 mol KCl/0.075L) = 9.85 M

Now that we have found the initial Molarity of the problem, we go back to the formula M(initial)V(initial)=M(final)V(final). And we are looking for Molarity(final) as shown in the question.

M(final)=M(initial)V(initial)/V(final)
M(final)=(9.85M)(0.075L)/(0.125L) = 5.91M (answer)


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