Module: Molarity and Dilution of a solution

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Cole Reynolds 4F
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Module: Molarity and Dilution of a solution

Postby Cole Reynolds 4F » Fri Oct 04, 2019 9:00 am

5.00 g of KMnO4 is dissolved in a 150.00 mL flask of water. If 20.00 mL of this solution is removed and placed in a new 2nd 250.00 mL flask and filled with water, what is the concentration of the solution in the 2nd flask? Molar Masses: K (39.10 g/mol), Mn (54.94 g/mol), O (16.00 g/mol).

I'm a bit confused with removing 20.00 mL of the solution and placing it in a new 2nd 250.00 mL flask.

Goyama_2A
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Re: Module: Molarity and Dilution of a solution

Postby Goyama_2A » Fri Oct 04, 2019 9:11 am

You need to calculate how many moles of KMnO4 you have using that 5.00 grams. Then, using that answer, calculate the molarity of the initial solution using the moles of KMnO4 divided by the 150 ml of water (but convert the volume to liters first). Then, using that molarity, you could use the M1V1 = M2V2 formula. M1 would be the initial molarity you just found and V1 would be the 20 ml (converted to liters) of the initial solution that you took out and transferred to the flask. The V2 would be the final volume (250 ml). You’d then have enough info to solve for your final molarity (M2), which would answer the question. Hope this helps!

ValerieChavarin 4F
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Re: Module: Molarity and Dilution of a solution

Postby ValerieChavarin 4F » Fri Oct 04, 2019 1:28 pm

This question requires the M1V1=M2V2 as it is a dilution problem. The 5.00g is necessary to find the moles of KMnO4. From there, you can calculate the M1, divide your answer by .15L (after unit conversion of mL). The 20.00mL or .02000L is your V1, with the new volume 250mL or .250L as your V2

Izamary Marquez 2L
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Re: Module: Molarity and Dilution of a solution

Postby Izamary Marquez 2L » Thu Oct 08, 2020 11:42 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?

A. 0.169 mol.L-1 B. 10.8 mol.L-1 C. 0.0923 mol.L-1 D. 5.91 mol.L-1



I'm a bit confused in the wording of this question...I used the M1V1=M2V2 equation and converted the g of KCl into moles first, solving for M2, but I keep getting 0.44 as my answer... Am I using the wrong equation/ taking the wrong approach?

Izamary Marquez 2L
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Joined: Wed Sep 30, 2020 9:44 pm

Re: Module: Molarity and Dilution of a solution

Postby Izamary Marquez 2L » Thu Oct 08, 2020 11:47 pm

Just a general question about the conversion of grams to moles and mL to L... I know we usually have to change the grams to moles as a first step to proceed with he problem, and with regards to molarity, we will likely be dealing with Liters.. However, I keep getting different answers due to the fact that sometimes I convert from mL to L after I solve the problem, and sometimes I convert it before..Is there a specific way we should go about this?

Chem_Mod
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Re: Module: Molarity and Dilution of a solution

Postby Chem_Mod » Fri Oct 09, 2020 2:37 pm

Hi Izamary, for your first question, you should look at the problem in terms of the final volume and final number of moles. Your total mass of KCl is 55.1 g and you are correct in first converting this to moles. Now to find the molarity, just divide by the total volume. The problem states that the final volume of water is 125 mL. Therefore, you would simply divide moles KCl by 0.125 L to get the molarity. You would not need to use M1V1=M2V2 in this problem.

For your second question about converting from mL to L, it should not matter when you do it. However, keep in mind that molarity M is ALWAYS moles per LITER. To help with conversions, try writing M as mol/L in your dimensional analysis equations instead of as M


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