[Win,odd Dhamnekar]

Calculate the ml that need to be transferred from a solution of 950ml CH₃COOH at 48% by mass with a density of 1.05g/ml to prepare a solution at 1.28M in 250ml.

My attempt:

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 [itex]152.51 ml=\frac{250 ml*2.098 M}{1.28 M}[/itex] needs to be transferred from 950 ml solution of CH₃COOH.

 Is my above answer correct?

[chenbeier]

I told you already  in the other forum.
NO.

[Orcio_87]

@Win,odd Dhamnekar

You are mistaking concentration (M = mol / litre) with quantity (mol).

[chenbeier]

Exactly