# 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.

###### 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:

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:

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)

ETHANOL

Given values:

Mass = 1 kg ethanol

Density of ethanol = 0.789 g/cm3

Calculate the volume of ethanol:

▪ convert 1 kg = 103 g

96% (485 ratings) ###### 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.