Solubility data of a certain solute with a certain solvent is empirical. There are constant values for this at varying temperatures. For KCl in water at 25°C, the solubility is 35.7 g/100 mL of water. When you compare this with the solubility data of KCl with ethanol, this means that KCl is more soluble in water than in ethanol. This is true because KCl is an ionic salt which is very soluble in water.
Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>
![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
=

= 0.001

The change in Concentration Δ
![[CH_{3}COOH]](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D)
= 0.001

CH3COOH H+ CH3COOH
Initial

0 0
Change

-0.001 +0.001 +0.001
Equilibrium

- 0.001 0.001 0.001
Since the

value is so small, the assumption
![[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D_%7Binitial%7D%20%3D%20%5BCH_%7B3%7DCOOH%5D_%7Bequilibrium%7D)
can be made.
![k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}](https://tex.z-dn.net/?f=%20k_%7Ba%7D%20%3D%20%5Btex%5D%3D%201.8%2A10%5E%7B-5%7D%20%20%3D%20%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%20%3D%20%20%5Cfrac%7B0.001%5E%7B2%7D%7D%7Bx%7D%20)
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!
In three hours and twenty minutes from 1:40 will be 5:00
A heterogeneous mixture is a mixture that it's components can be easily seen and separated.
c. a jar of mixed nuts