Module: Molarity and Dilution of a solution

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Cole Reynolds 4F
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Joined: Thu Jul 11, 2019 12:17 am

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
Posts: 107
Joined: Sat Aug 24, 2019 12:17 am

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
Posts: 99
Joined: Wed Sep 18, 2019 12:18 am

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


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