post assessment

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Maggie Doan 1I
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

post assessment

Postby Maggie Doan 1I » Sat Oct 06, 2018 1:00 am

What volume of 0.0380 M KMnO4 is needed to prepare 250 mL of 1.50 x 10-3 M KMnO4?
Would you just need to set them equal to each other?

yast_27
Posts: 29
Joined: Fri Sep 28, 2018 12:19 am

Re: post assessment

Postby yast_27 » Sat Oct 06, 2018 9:08 am

This is a dilution problem, typicaly you have to use the following equality:

C1.V1 = C2.V2

Here you have:

C1 = 0.0380 = 3.80 x 10^-2
V1 = Unknown
C2 = 1.50 x 10^-3
V2 = 250mL = 0.25 x 10^-3L

You have 1 value that is unknown so you just transform the equality so that you end up with

V1 = (C2.V2)/C1

and you just replace with the value that you have to find the answer.

I hope that this will help!


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