Calculate Molarity/Mole Fraction of Acetone in 1.00 M Solution

[lha08]

Homework Statement

Calculate the molarity and mole fraction of acetone in a 1.00 m solution of acetone, CH3COCH3 (d=0.788 g/mL) in ethanol, C2H5OH (d=0.789 g/mL). Assume that the volumes of acetone and ethanol add.

Homework Equations

The Attempt at a Solution

I'm not sure if I'm doing the question correctly but here it goes:
1. I first assumed 1 kg of Ethanol, C2H5OH.
2. With this, I used the 1.00 m and found 1.00 mol of acetone. CH3COCH3.
3. With the 1.00 mol acetone i found the number of grams using the molar mass. (58.09 g acetone)
4. I used the density of acetone and found the volume of acetone (73.72 mL)
5. Since 1 kg ethanol is 1000 g ethanol, i found the volume also by using the density. (1.27L)
6. I added both volumes together in Liters (1.3437 L)
7. Found the Molarity (1.00 mol/1.3437 L) and the Mol Fraction (1 mol/(1 + 21.70 mol)

Is this right?? Thanks

 

[symbolipoint]

Your method for molarity calculation is good.

 

[Blueshift5]

for your help!

Your approach seems to be correct. However, there are a few things to note:

1. The molarity of the solution is actually 1.00 M, not 1.00 m. The lowercase "m" is used for molality, which is a different unit of concentration that takes into account the mass of the solvent. Molarity, on the other hand, is the number of moles of solute per liter of solution.

2. When calculating the volume of acetone, you should use the density of acetone (0.788 g/mL) and the mass of acetone (58.09 g), not the molar mass. This is because the molar mass is the mass of one mole of acetone, which is different from the mass of the amount of acetone in the solution (58.09 g).

3. Similarly, when calculating the volume of ethanol, you should use the density of ethanol (0.789 g/mL) and the mass of ethanol (1000 g), not the molar mass. This is because the molar mass is the mass of one mole of ethanol, which is different from the mass of the amount of ethanol in the solution (1000 g).

4. When calculating the mole fraction, you should use the moles of acetone (1.00 mol) and the total moles of solute (1.00 mol + 21.70 mol), not the moles of acetone and the total volume of solution.

So the correct calculations would be:

1. Mass of acetone = 1.00 mol x 58.09 g/mol = 58.09 g

2. Volume of acetone = 58.09 g / 0.788 g/mL = 73.64 mL

3. Volume of ethanol = 1000 g / 0.789 g/mL = 1269 mL = 1.269 L

4. Total volume of solution = 73.64 mL + 1.269 L = 1.342 L

5. Molarity of acetone = 1.00 mol / 1.342 L = 0.745 M

6. Mole fraction of acetone = 1.00 mol / (1.00 mol + 21.70 mol) = 0.0448

So the molarity of acetone in the solution is 0.745 M