posted by Mitch .
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.