# Chemistry

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Calculate the molality (m) of a 6.56 kg sample of a solution of the solute acetone dissolved in the solvent ethanol if the sample contains 1.44 kg of acetone.

Molar Mass (g/mol)
acetone = 58.05
ethanol = 46.07

Density (g/mL)
acetone = 0.7899
ethanol = 0.7893

Dr. Bob, I asked a question like this about molality and the step by step explanation that you provided helped greatly. Could you please help me out with this one. It is slightly different and I'm really confused. Thank you!

• Chemistry -

never mind, I figured it out. I'll post the explanation just in case anyone else needs help with a similar problem.

Equation:

molality acetone = mol acetone/kg ethanol
-------------------------------------
to find kg ethanol:

total mass - sample mass of acetone
(6.56kg - 1.44kg) = 5.12kg ethanol
-------------------------------------
to find mol acetone:

convert kg to g of sample mass given
(1.44kg x 1000g/kg) = 1440g acetone

divide sample mass(grams) by molar mass
(1440g / 58.05g)= 24.80620155mol acetone
-------------------------------------

Molality = 24.80620155 mol acetone/ 5.12kg ethanol

• Chemistry -

I didn't go through the math but this looks ok to me.

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