[lha08]

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.

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

[Borek]

This is a correct approach, I have not checked the numbers.