Process of Molarity Calculation

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Saumya Tawakley 1E
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Process of Molarity Calculation

Postby Saumya Tawakley 1E » Mon Oct 05, 2020 11:55 am

In the Molarity and Dilution of a Solution: Post-Module Assessment, there was a question:

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?

For solving this problem, could we find the molarity by converting 55.1g KCl to moles and dividing directly by 0.125L, or would we need to find the initial molarity in 75mL of water then use M1V1 = M2V2? In this problem directly dividing by 0.125L seemed to work, but would that process work correctly for every problem of this type or would we need to use M1V1 = M2V2?

Thanks!

Steph Du 1H
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Re: Process of Molarity Calculation

Postby Steph Du 1H » Mon Oct 05, 2020 11:57 am

Both methods work, it's just that sometimes using M1V1=M2V2 to solve it will be a lot faster than your first method. Depends on the question really but both are correct.

EnricoArambulo3H
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Re: Process of Molarity Calculation

Postby EnricoArambulo3H » Mon Oct 05, 2020 11:58 am

I think you could use the dilution equation since you are being given two different volumes. The solution in 75 mL of water will have a different concentration than the solution in 125 mL. However since moles are constant regardless of volume, you can also use M=n/V.
Last edited by EnricoArambulo3H on Mon Oct 05, 2020 1:03 pm, edited 1 time in total.

Stuti Pradhan 2J
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Re: Process of Molarity Calculation

Postby Stuti Pradhan 2J » Mon Oct 05, 2020 11:59 am

Since the moles of KCL are the same throughout the entire process and all the water is retained (50 mL are added to the initial 75mL, no portion of the 75 mL are transferred elsewhere), you can divide directly by 0.125 L to find the molarity.

Hope this helps!

Margia Adriano 2A
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Re: Process of Molarity Calculation

Postby Margia Adriano 2A » Mon Oct 05, 2020 12:12 pm

Your first method and your second method are one in the same in this case because the first method just skips a step in setting up the second. M*V is equal to the moles of solute because Molarity is equal to moles/volume, so when you move these parts around, moles is also equal to volume times molarity. So if we were to set it up with MV=MV, it would have been (9.85)(.075)=M*(.125). When you multiply out the (9.85)(.075), you get 0.739 moles of solute, which you had already solved for earlier. To finish this equation, you would just divide 0.739/0.125, which is the same answer as if you were to skip setting up the whole thing with MV=MV. So really, your first method would always work for this type of question.

Ria Nawathe 1C
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Re: Process of Molarity Calculation

Postby Ria Nawathe 1C » Mon Oct 05, 2020 12:14 pm

You would use the first method and directly divide the moles of KCl by the final volume since the moles of solute remain constant through the process, as a previous answer mentioned. In terms of when to use the M1V1 = M2V2 equation, you would typically use it for problems where you have to calculate the volume of stock solution to dilute. Ex: How much X M stock solution do you need to prepare Y L of a Z M solution?

Sahaj Patel Lec3DisK
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Re: Process of Molarity Calculation

Postby Sahaj Patel Lec3DisK » Mon Oct 05, 2020 3:19 pm

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?

You would first convert 55.1 G of KCl to MOLS, then convert 75 mL to Liters. You would then divide the MOLS over Liters to get the Molarity of that specific solution. From there, you simply use the dilution equation to find the MOLS of KCl in the new solution. Afterwards, you would divide the MOLS of the new KCl solution by the 0.125 L to find the answer.


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