Chemistry Forum

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

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.

mmmmm thank you ?

What is wrong, what you dont understand. Start with the first one.
What is meant with 1.28 M

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.

Wrong result

I edited my answer from 152.5 ml to 145.12 ml. Is this correct?

No, show your calculation. But normaly Sue should do it.

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

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.

Thanks for your correct answer.