Serial Dilutions - AS Biology

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anona1255
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#1
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#1
Hi, I was wondering whether there was a formula to calculate the end concentrations of serial dilutions and how I would apply this to questions.

E.g. 1cm^3 of 2M hydrochloric acid was added to 9cm^3 of distilled water. The mixture was inverted, then 1cm^3 of the mixture was added to a further 9cm^3 of distilled water. What was the final concentration?

Thanks in advance.
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Jsisnxjdnd
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#2
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C1*v1=c2*v2
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aytuiq
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#3
0.02moldm^-3
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anona1255
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#4
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(Original post by Jsisnxjdnd)
C1*v1=c2*v2
Ah thank you, but how would you apply this to this question in particular?
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anona1255
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(Original post by aytuiq)
0.02moldm^-3
Thank you
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aytuiq
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#6
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No problem !

Good luck with your exam !
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aytuiq
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Explanation to answer :

1cm^3 2M HCl added to 9cm^3 of water means there is a final volume of 10cm^3 with a dilution of 1:10 and 2M/10=0.2M,
This is repeated with 9cm^3 again, so divide concentration by 10 again.


Further help:

https://socratic.org/questions/how-d...n-calculations
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anona1255
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#8
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(Original post by aytuiq)
Explanation to answer :

1cm^3 2M HCl added to 9cm^3 of water means there is a final volume of 10cm^3 with a dilution of 1:10 and 2M/10=0.2M,
This is repeated with 9cm^3 again, so divide concentration by 10 again.


Further help:

https://socratic.org/questions/how-d...n-calculations
Thank you so much! You too
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Namih
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#9
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#9
(Original post by anona1255)
Ah thank you, but how would you apply this to this question in particular?
It applies to all dilution calculations. You want to keep the moles before the same as moles after (moles being CV)
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