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July 26th, 2017, 05:10 AM   #1
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What volume of water should be added?

What volume of water should be added to 40.0mL acetic acid B.P. (33% w/w, density of 1.041g/mL) such that the final concentration of acetic acid in the solution is 10% w/v? (The molecular weight of acetic acid is 60.0). The answer is 97.4mL

I've worked out two ways to get the answer, but I'm not sure whether the methods are correct.

Solution 1:
33% w/w = 33g acetic acid per 100g solution
100g / (1.041g/mL) = 96.06mL solution
Thus, 33g acetic acid per 96.06mL solution [1]
X g acetic acid per 40.0mL solution [2]
Using cross multiplication, I worked out x using [1] and [2]
40x33=96.06x
X=13.74g
Thus, 13.74g acetic acid in 40.0mL solution
(13.74g)/(40mL+xmL) x 100 = 10%w/v
Working that out, I got x=97.4mL

Solution 2:
33%w/w x 1.041g/mL = 34%w/v
C1 = 34%w/v, V1= 40mL, C2= 10% w/v, V2=?
Using C1V1=C2V2, I got V2=97.4mL

Any input would be appreciated. =)

Last edited by skipjack; July 26th, 2017 at 08:25 AM.
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July 26th, 2017, 09:10 AM   #2
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What does "w/v" in the original description mean?

How exactly did you get 97.4mL in your solution 2?
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July 26th, 2017, 02:54 PM   #3
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Sorry, I should have named solution 1 and 2 as method 1 and method 2 instead. I just worked out 2 ways to find the answer, but I don't know whether it's the right approach.

Last edited by skipjack; July 26th, 2017 at 03:36 PM.
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July 26th, 2017, 03:38 PM   #4
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How did you get 97.4mL using your second method? Can you give the details of your arithmetic?
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July 26th, 2017, 06:50 PM   #5
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33%w/w x 1.041g/mL
=33g/100g x 1.041g/mL = 34%w/v
C1 = 34%w/v, V1= 40mL, C2= 10% w/v, V2=?
Using C1V1=C2V2, I got V2=97.4mL


Just looking at my working out above, it doesn't look like the calculation for lines 1 and 2 are possible. Maybe method 2 isn't the correct way to solve this problem?
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July 27th, 2017, 08:45 AM   #6
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Quote:
Originally Posted by pianist View Post
Using C1V1=C2V2, I got V2=97.4mL
What are the full details of the arithmetic you did for this step? Typing "I got" doesn't tell me what exactly you did. Also, you have at all stages omitted to say when you have rounded your calculated values and what the unrounded values were.

Have you checked whether "w/v" was a typo?
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July 27th, 2017, 05:08 PM   #7
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The w/v in the question isn't a typo.

33%w/w x 1.041g/mL
=33g/100g x 1.041g/mL = 34.353%w/v
C1 = 34.353%w/v, V1= 40mL, C2= 10% w/v, V2=?
Using C1V1=C2V2,

(34.353)(40)=(10)(V2)
V2=137.412mL

137.412mL - 40mL = 97.412mL of water needs to be added
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