Determine the molarity and mole fraction of a 1.15 m solution of acetone (CH3COCH3) dissolved in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the final volume equals the sum of the volumes of acetone and ethanol.

The molarity is 0.85M

The mole fraction = 0.05

Explanation:

Step 1: Data given

Molarity of acetone = 1.15 m

Density of acetone = 0.788 g/cm³

Density of ethanol = 0.789 g/cm³

Molar mass of acetone = 58.08 g/mol

Step 2: Calculate number of moles acetone

Molality = moles solute / kg solvent = moles acetone / kg ethanol. (To make it easy, we will suppose we have 1 kg ethanol)

1.15 m = 1.15 moles acetone / 1 kg ethanol

Step 3: Calculate mass of acetone

1.15 moles acetone * (58.08 g/mol) = 66.792 g acetone

Step 4: Calculate volume of acetone

66.792 g acetone / 0.788 g/mL acetone = 84.76 mL acetone

Step 5: Calculate volume of ethanol

1000 g ethanol / 0.789 g/mL ethanol = 1267.43 mL ethanol

Step 6: Calculate total volume solution

Total solution volume = 84.76 + 1267.43 = 1352.2 mL = 1.3522L

Step 7: Calculate molarity

Molarity of acetone = moles acetone / volume solution = 1.15moles / 1.3522L

Molarity = 0.85 M

Step 8: Calculate moles ethanol

moles ethanol = mass / molar mass = 1000g / 46.0g/mol = 21.74 moles

Step 9: Calculate mole fraction

mole fraction acetone = (moles acetone / total moles) = (1.15 / (1.15 + 21.74)) = 0.05