How to make 1.5 molar solution of Acetic Acid and Toluene?

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Alexander Kolker

Home-used acetic acid is an aqueous solution 5% or 9%. It is not mixed well with toluene (water is practically insoluble in toluene).

Therefore you have to take a glacial acetic acid which is water-free (anhydrous) acetic acid but practically contains a trace of water, usually it is 99.5% pure acetic acid. MW=60.05 g/mol.

In order to make 1.5M solution in toluene you have to take 1.5M*60.05g =90.075 g pure glacial acetic acid or 90.075/0.995=90.527 g of the commercial glacial acid; dissolve that amount in 0.5 L of toluene, and then make the total volume up to exactly 1 L by adding more toluene. Note, you cannot dissolve it at once in 1 L of toluene because the volume upon mixing is changed.

Carlos Araújo Queiroz

You may want to consider liquid-liquid equilibrium data for the water/acetic acid/toluene system; cf.:

Article Liquid–Liquid Equilibrium Data, Density, Viscosity, and Inte...

Article (Liquid + liquid) equilibria for the ternary (water + acetic...

Article THE DISTRIBUTION COEFFICIENTS OF ACETIC ACID BETWEEN WATER A...

https://www.jstage.jst.go.jp/article/jcej1968/12/3/12_3_236/_article/-char/en