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.

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:

$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\frac{\mathbf{mol}}{\mathbf{kg}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{moles}\mathbf{ }\mathbf{acetone}}{\mathbf{1}\mathbf{ }\mathbf{kg}}$

$\left(\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\frac{\mathbf{mol}}{\sout{\mathbf{kg}}}\right)\mathbf{\times }\mathbf{1}\mathbf{ }\sout{\mathbf{kg}}\mathbf{ }\mathbf{=}\mathbf{ }\left(\frac{\mathbf{moles}\mathbf{ }\mathbf{acetone}}{\sout{\mathbf{1}\mathbf{ }\mathbf{kg}}}\right)\mathbf{\times }\sout{\mathbf{1}\mathbf{ }\mathbf{kg}}$

$\mathbf{moles}\mathbf{ }\mathbf{acetone}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{acetone}$

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:

$\mathbf{mass}\mathbf{ }\mathbf{acetone}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\sout{\mathbf{mol}\mathbf{ }\mathbf{acetone}}\mathbf{\times }\frac{\mathbf{58}\mathbf{.}\mathbf{078}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{acetone}}{\mathbf{1}\mathbf{ }\sout{\mathbf{mol}\mathbf{ }\mathbf{acetone}}}$

$\mathbf{mass}\mathbf{ }\mathbf{acetone}\mathbf{ }\mathbf{=}\mathbf{58}\mathbf{.}\mathbf{078}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{acetone}$

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)

${\mathbf{V}}_{\mathbf{acetone}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{58}\mathbf{.}\mathbf{078}\mathbf{ }\sout{\mathbf{g}}\mathbf{ }\mathbf{acetone}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\sout{{\mathbf{cm}}^{\mathbf{3}}}}{\mathbf{0}\mathbf{.}\mathbf{788}\mathbf{ }\sout{\mathbf{g}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\sout{\mathbf{mL}}}{\mathbf{1}\mathbf{ }\sout{{\mathbf{cm}}^{\mathbf{3}}}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{L}}{\mathbf{1}\mathbf{ }\sout{\mathbf{mL}}}$

${\mathbf{V}}_{\mathbf{acetone}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{0737}\mathbf{ }\mathbf{L}\mathbf{ }\mathbf{acetone}$

ETHANOL

Given values:

Mass = 1 kg ethanol

Density of ethanol = 0.789 g/cm³

Calculate the volume of ethanol:

▪ convert 1 kg = 10³ g

${\mathbf{V}}_{\mathbf{e}\mathbf{t}\mathbf{h}\mathbf{a}\mathbf{n}\mathbf{o}\mathbf{l}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{ }\sout{\mathbf{k}\mathbf{g}}\mathbf{ }\mathbf{ethanol}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{3}}\mathbf{ }\sout{\mathbf{g}}}{\mathbf{1}\mathbf{ }\sout{\mathbf{kg}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\sout{{\mathbf{cm}}^{\mathbf{3}}}}{\mathbf{0}\mathbf{.}\mathbf{789}\mathbf{ }\sout{\mathbf{g}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\sout{\mathbf{mL}}}{\mathbf{1}\mathbf{ }\sout{{\mathbf{cm}}^{\mathbf{3}}}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{L}}{\mathbf{1}\mathbf{ }\sout{\mathbf{mL}}}$

${\mathbf{V}}_{\mathbf{ethanol}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{2674}\mathbf{ }\mathbf{L}\mathbf{ }\mathbf{ethanol}$