acetic acid solution
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acetic acid solution
Calculate the ml that need to be transferred from a solution of 950ml CH 3 COOH at 48% by mass with a density of 1.05g / ml to prepare a solution at 1.28M in 250ml.
i dont understand
i dont understand
- ChenBeier
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Re: acetic acid calculation
Calculate backward how many moles is 1,28 M in 250 ml, this convert with the molar mass of acetic acid to the mass. This value dilute to 48 % and with the density you get the required volume.
Re: acetic acid solution
mmmmm thank you ?
- ChenBeier
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Re: acetic acid solution
What is wrong, what you dont understand. Start with the first one.
What is meant with 1.28 M
What is meant with 1.28 M
- Dhamnekar Winod
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Re: acetic acid solution
If i am correct , 145.12 ml that need to be transferred from the solution of 950 ml of CH3COOH at 48% by mass with a density of 1.05 g/ml to prepare solution of 1.28 M in 250 ml.
Last edited by Dhamnekar Winod on Thu Mar 25, 2021 5:11 am, edited 1 time in total.
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- ChenBeier
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Re: acetic acid solution
Wrong result
- Dhamnekar Winod
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Re: acetic acid solution
I edited my answer from 152.5 ml to 145.12 ml. Is this correct?
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- ChenBeier
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Re: acetic acid solution
No, show your calculation. But normaly Sue should do it.
- Dhamnekar Winod
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Re: acetic acid solution
Here is my computational chemistry work
We have 950 ml of solution of CH3COOH.
The density of this solution is 1.05g/ml. CH3COOH is present 48% by mass in this solution. Mass of 950 ml of solution is 950 ml *1.05g= 997.50 g.
CH3COOH is 0.48 of 997.50 g. = 478.80 g. Molar mass of CH3COOH is 60.0524g/gmol. 478.8g/60.0524 g = 7.973 M of CH3COOH is present in 950 l of solution.
So, in 250 ml of its solution ,2.098 M of CH3COOH is present.
But we want to prepare solution of 1.28 M. Hence, we require \(152.51 ml =\frac{ 250 ml*2.098M}{1.28 M}\) needs to betransferred from 950 ml of solution of CH3COOH
We have 950 ml of solution of CH3COOH.
The density of this solution is 1.05g/ml. CH3COOH is present 48% by mass in this solution. Mass of 950 ml of solution is 950 ml *1.05g= 997.50 g.
CH3COOH is 0.48 of 997.50 g. = 478.80 g. Molar mass of CH3COOH is 60.0524g/gmol. 478.8g/60.0524 g = 7.973 M of CH3COOH is present in 950 l of solution.
So, in 250 ml of its solution ,2.098 M of CH3COOH is present.
But we want to prepare solution of 1.28 M. Hence, we require \(152.51 ml =\frac{ 250 ml*2.098M}{1.28 M}\) needs to betransferred from 950 ml of solution of CH3COOH
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
- ChenBeier
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Re: acetic acid solution
No that was not the question.
A solution of 250 ml 1.28 M should be prepared.
250 ml contain 1/4 = 0.32 mol. This calculated with the molar mass of 60 g/mol gives 19,2 g 100% what is equal 40 g 48%. With Density of 1.05 g/ ml it gives 38 ml.
A solution of 250 ml 1.28 M should be prepared.
250 ml contain 1/4 = 0.32 mol. This calculated with the molar mass of 60 g/mol gives 19,2 g 100% what is equal 40 g 48%. With Density of 1.05 g/ ml it gives 38 ml.
- Dhamnekar Winod
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Re: acetic acid solution
Thanks for your correct answer.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]