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Chapter 10: Problem 2

Acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\), is the main ingredient of nail polish remover. A solution is made up by adding \(35.0 \mathrm{~mL}\) of acetone \((d=0.790 \mathrm{~g} / \mathrm{mL})\) to \(50.0 \mathrm{~mL}\) of ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(d=0.789 \mathrm{~g} / \mathrm{mL})\). Assuming volumes are additive, calculate (a) the mass percent of acetone in the solution. (b) the volume percent of ethyl alcohol in the solution. (c) the mole fraction of acetone in the solution.

Short Answer

Expert verified
Answer: (a) Mass percent of acetone: 41.2% (b) Volume percent of ethyl alcohol: 58.8% (c) Mole fraction of acetone: 0.358

Step by step solution

01

Find the mass of acetone and ethyl alcohol

First, find the mass of acetone by multiplying its volume with its density: Mass of acetone = \(\mathrm{Volume} \times \mathrm{Density} = 35.0 \mathrm{~mL} \times 0.790~\mathrm{g/mL} = 27.65\mathrm{~g}\) Next, find the mass of ethyl alcohol by multiplying its volume with its density: Mass of ethyl alcohol = \(\mathrm{Volume} \times \mathrm{Density} = 50.0 \mathrm{~mL} \times 0.789~\mathrm{g/mL} = 39.45\mathrm{~g}\)
02

Find the total mass and volume of the solution

Now, find the total mass of the solution by adding the mass of acetone and ethyl alcohol: Total mass of solution = Mass of acetone + Mass of ethyl alcohol = \(27.65\mathrm{~g} + 39.45\mathrm{~g} = 67.10\mathrm{~g}\) The total volume of the solution is given as the sum of volumes of acetone and ethyl alcohol: Total volume of solution = \(35.0\mathrm{~mL} + 50.0\mathrm{~mL} = 85.0\mathrm{~mL}\)
03

Calculate the mass percent of acetone in the solution

Using the mass of acetone and the total mass of the solution, we can calculate the mass percent of acetone: Mass percent of acetone = \(\frac{\text{Mass of acetone}}{\text{Total mass of solution}} \times 100\% = \frac{27.65\mathrm{~g}}{67.10\mathrm{~g}} \times 100\% = 41.2\%\)
04

Calculate the volume percent of ethyl alcohol in the solution

Using the volume of ethyl alcohol and the total volume of the solution, we can calculate the volume percent of ethyl alcohol: Volume percent of ethyl alcohol = \(\frac{\text{Volume of ethyl alcohol}}{\text{Total volume of solution}} \times 100\% = \frac{50.0\mathrm{~mL}}{85.0\mathrm{~mL}} \times 100\% = 58.8\%\)
05

Calculate the moles of acetone and ethyl alcohol

To find the mole fraction of acetone, we first need to find the moles of acetone and ethyl alcohol in the solution. Moles of acetone = \(\frac{\text{Mass of acetone}}{\text{Molar mass of acetone}} = \frac{27.65\mathrm{~g}}{58.08\mathrm{~g/mol}} = 0.476\mathrm{~mol}\) Moles of ethyl alcohol = \(\frac{\text{Mass of ethyl alcohol}}{\text{Molar mass of ethyl alcohol}} = \frac{39.45\mathrm{~g}}{46.07\mathrm{~g/mol}} = 0.856\mathrm{~mol}\)
06

Calculate the mole fraction of acetone in the solution

Now we can calculate the mole fraction of acetone: Mole fraction of acetone = \(\frac{\text{Moles of acetone}}{\text{Total moles}} = \frac{0.476\mathrm{~mol}}{0.476\mathrm{~mol} + 0.856\mathrm{~mol}} = 0.358\) The answers are: (a) mass percent of acetone: \(41.2\%\) (b) volume percent of ethyl alcohol: \(58.8\%\) (c) mole fraction of acetone: \(0.358\)

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