A solution is prepared by dissolving 55.1g of KCl in approximately 75mL of water and then adding water to a final volume of 125mL. What is the molarity of KCl(aq) in this solution?
I know that Molarity equals:
Molarity=moles of solute/ volume of the solution
but in this case would we not use this formula rather the other formula Professor Lavelle used in class of :
Mintial x Vinitial= Mfinal x Vfinal
and then reworking that equation to solve for the Molarity?
Molarity and Dilution [ENDORSED]
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Re: Molarity and Dilution [ENDORSED]
In this case it is possible to use the MiVi = MfVf equation. Mi = 9.85M and Vi = .075L and Vf = .125L. So, if you multiply Mi by Vi then divide this product by Vf you end up with 5.91M. Another way you could solve this problem is simply converting the grams into moles, which would be .739 mol, then dividing this by the final volume (.125L) to get 5.91M as well since the moles do not change since KCl is neither added nor removed.
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Re: Molarity and Dilution
You can still use the Molarity = mol/vol equation, the question is just worded in an unusual way. I could be wrong, but in summary I think the problem was just stating that 55.1g of KCl is added to 125 mL of water.
Re: Molarity and Dilution
A solution is prepared by dissolving 55.1g of KCl in approximately 75mL of water and then adding water to a final volume of 125mL. What is the molarity of KCl(aq) in this solution?
I know that Molarity equals:
Molarity=moles of solute/ volume of the solution
but in this case would we not use this formula rather the other formula Professor Lavelle used in class of :
Mintial x Vinitial= Mfinal x Vfinal
and then reworking that equation to solve for the Molarity?
1. Convert 55.1g of KCl into moles by dividing by the molar mass of KCl (74.55 g/mol) = 0.7391 mol KCl
2. Convert mol KCl to molarity using formula (M=n/V); convert volume to liters... M=0.7391mol/.075L = 9.85M
3. Plug in values into M1V1=M2V2 formula = (9.85M)(.075L)=(M2)(.125L)
4. M2 = 5.91 (watch sig figs)
I know that Molarity equals:
Molarity=moles of solute/ volume of the solution
but in this case would we not use this formula rather the other formula Professor Lavelle used in class of :
Mintial x Vinitial= Mfinal x Vfinal
and then reworking that equation to solve for the Molarity?
1. Convert 55.1g of KCl into moles by dividing by the molar mass of KCl (74.55 g/mol) = 0.7391 mol KCl
2. Convert mol KCl to molarity using formula (M=n/V); convert volume to liters... M=0.7391mol/.075L = 9.85M
3. Plug in values into M1V1=M2V2 formula = (9.85M)(.075L)=(M2)(.125L)
4. M2 = 5.91 (watch sig figs)
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Re: Molarity and Dilution
You do not have to do M1xV1=M2V2 because the question is asking for molarity and not dilution. You still can but it will take you longer.
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Re: Molarity and Dilution
You could just use M = n/V, you just have to convert the grams of KCl to moles, and then use .125L as your final volume.
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Re: Molarity and Dilution
I have a hard time identifying what is what in wordy questions. For instance I often mix up what is the Molarity, number of moles of solute, or the volume. And I have a hard time with excluding unnecessary givens in the equation. Does anyone have any advice?
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Re: Molarity and Dilution
Jennah Muhammad 1H wrote:I have a hard time identifying what is what in wordy questions. For instance I often mix up what is the Molarity, number of moles of solute, or the volume. And I have a hard time with excluding unnecessary givens in the equation. Does anyone have any advice?
Molarity is moles/Liters and is sometimes described as M. Moles are usually written out as mol or n. Volume is anything in liters,L. I guess it's more of a memorization thing.
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