Problem: Calculate the molarity and mole fraction of acetone in a 1.00 m solution of acetone (CH3COCH3) in ethanol (C 2H5OH). (Density of acetone = 0.788 g/cm 3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

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FREE Expert Solution

Step 1. Determine the moles of acetone in the solution.

We are given a molality of 1.00 m solution of acetone in ethanol

Assume that there is 1 kg of solvent ethanol:


1.00 molkg = moles acetone1 kg


1.00 molkg×1 kg = moles acetone1 kg×1 kg


moles acetone = 1.00 mol acetone



Step 2. Calculate the mass of acetone (in g) of the solution

We can calculate the mass of acetone using molar mass.

The molar mass of acetone (CH3COCH3) is:

3 C x 12.01 g/mol C = 36.03 g/mol

6 H x 1.008 g/mol H = 6.048 g/mol

1 O x 16.00 g/mol O = 16.00 g/mol

Molar mass = 58.078 g/mol CH3OHCH3


Calculate the mass of acetone in kg:


mass acetone = 1.00 mol acetone×58.078 g acetone1 mol acetone


mass acetone =58.078 g acetone



Step 3. Calculate the volume of ethanol (in L) present in the solution.

We can calculate the volume by using mass and density:

Mass (density) → volume



ACETONE

Given values:

Mass = 69.694 g acetone

Density of acetone = 0.788 g/cm3


The volume in the density given is in cm3, we will have to convert it to the required unit in L.

Conversion units:    1 mL = 1 cm3

  1 mL = 10-3 L


Calculate the volume of acetone:

▪ Put units you need to cancel out on opposite places (if it’s in the numerator then put the conversion factor with the same unit in the denominator)


Vacetone = 58.078 g acetone×1 cm30.788 g×1 mL1 cm3×10-3 L1 mL


Vacetone = 0.0737 L acetone



ETHANOL

Given values:

Mass = 1 kg ethanol

Density of ethanol = 0.789 g/cm3


Calculate the volume of ethanol:

▪ convert 1 kg = 103 g


Vethanol = 1 kg ethanol×103 g1 kg×1 cm30.789 g×1 mL1 cm3×10-3 L1 mL


Vethanol = 1.2674 L ethanol


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Problem Details

Calculate the molarity and mole fraction of acetone in a 1.00 m solution of acetone (CH3COCH3) in ethanol (C 2H5OH). (Density of acetone = 0.788 g/cm 3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.