Calculate the volume ( mL ) of the solute acetic acid and the volume ( L ) of the solvent H2O that should be added to generate 5.01 kg of a solution that is 0.694 m acetic acid.

volume ( mL ) of the CH3CO2H

volume ( L ) of the H2O

i just want to check my answers...its either i get 100% or 60% if i get this question right.

i got 191.2 as the frist one then 4.82 as the second... any confirmation?

What are you using for the density of acetic acid and water?

1.049 and .9982 respectively.

I don't confirm that.

First, I am using 1.049 g/mL for the density of acetic acid and 1.00 g/mL for the density of water.
191.2 mL acetic acid is 191.2/1.049 = 200.57 g and that is 200.57/60.05 = 3.34 moles acetic acid.
3.34/4.82 kg = 0.69295 and round to 0.693 which isn't 0.694.

2. There is a second problem.
If we take the 3.34 moles acetic acid which has a mass of 200.57 g and add it to 4820 g H2O we end up with a solution mass of 4820+200.57 = 5020.6 or 5121 rounded or 5.12 kg and not 5.01 kg.

The 0.9982 may make a difference. I used 1.00.

Let me see.

That's closer but not quite what I obtained.

191.2 mL acetic acid = 200.57 grams

4.82 L = 4820 mL
4820 x 0.9982 = 4811.3 grams
Total mass of solution =
4811.3 + 200.57 = 5011.9 = 5.0119 kg
3.34/4.8113 = 0.694
So m is right there. The total mass is just a hair over 5.01 kg. I used a mass of 4809.57 g H2O (or 4809.6 g) which is about 1 mL less than your calculations.

To calculate the volume of solute (acetic acid) and the volume of solvent (water) needed to generate the desired solution, we need to use the formula:

Molarity = Moles of Solute / Volume of Solution (in liters)

First, let's calculate the moles of acetic acid required:

Molarity = 0.694 m = 0.694 moles of acetic acid / 1 L of solution

Next, we'll calculate the moles of acetic acid needed to generate 5.01 kg of solution:

Mass of acetic acid = 5.01 kg
Molar mass of acetic acid (CH3CO2H) = 60.05 g/mol

Moles of acetic acid = (Mass of acetic acid / Molar mass of acetic acid)
= (5.01 kg / 60.05 g/mol)
= 83.212 mol

Now let's calculate the volume of acetic acid:

Volume of acetic acid (in mL) = (Moles of acetic acid / Molarity) * 1000
= (83.212 mol / 0.694 mol/L) * 1000
= 119.83 mL
≈ 120 mL

Lastly, let's calculate the volume of water:

Volume of water (in L) = Volume of Solution - Volume of acetic acid
= 5 L - 0.120 L
= 4.88 L
≈ 4.82 L

So, based on the calculations, the volume of solute (acetic acid) should be approximately 120 mL, and the volume of solvent (water) should be approximately 4.82 L.

Your answers of 191.2 mL for acetic acid and 4.82 L for water appear to be incorrect.