Acetone is the main ingredient of nail polish remover. A solution is made up by adding 35.0 ml acetone (density = 0.790 g/ml) to 50.0 ml of ethyl alcohol,  (density =0.789 g/ml), Assuming volumes are additive. Calculate, Mass percent of acetone, the Volume percent of ethyl alcohol, and Mole fraction of acetone
The Correct Answer and Explanation is :
Let’s solve for the mass percent of acetone, the volume percent of ethyl alcohol, and the mole fraction of acetone in the solution.
1. Mass Percent of Acetone
The mass percent of acetone is calculated as:
[
\text{Mass percent of acetone} = \frac{\text{Mass of acetone}}{\text{Total mass of solution}} \times 100
]
Step 1: Calculate the mass of acetone
- Volume of acetone = 35.0 mL
- Density of acetone = 0.790 g/mL
[
\text{Mass of acetone} = \text{Volume} \times \text{Density} = 35.0 \, \text{mL} \times 0.790 \, \text{g/mL} = 27.65 \, \text{g}
]
Step 2: Calculate the mass of ethyl alcohol
- Volume of ethyl alcohol = 50.0 mL
- Density of ethyl alcohol = 0.789 g/mL
[
\text{Mass of ethyl alcohol} = \text{Volume} \times \text{Density} = 50.0 \, \text{mL} \times 0.789 \, \text{g/mL} = 39.45 \, \text{g}
]
Step 3: Calculate the total mass of the solution
[
\text{Total mass} = \text{Mass of acetone} + \text{Mass of ethyl alcohol} = 27.65 \, \text{g} + 39.45 \, \text{g} = 67.10 \, \text{g}
]
Step 4: Calculate the mass percent of acetone
[
\text{Mass percent of acetone} = \frac{27.65 \, \text{g}}{67.10 \, \text{g}} \times 100 = 41.2\%
]
2. Volume Percent of Ethyl Alcohol
The volume percent of ethyl alcohol is calculated as:
[
\text{Volume percent of ethyl alcohol} = \frac{\text{Volume of ethyl alcohol}}{\text{Total volume of solution}} \times 100
]
Step 1: Calculate the total volume of the solution
- Volume of acetone = 35.0 mL
- Volume of ethyl alcohol = 50.0 mL
[
\text{Total volume} = 35.0 \, \text{mL} + 50.0 \, \text{mL} = 85.0 \, \text{mL}
]
Step 2: Calculate the volume percent of ethyl alcohol
[
\text{Volume percent of ethyl alcohol} = \frac{50.0 \, \text{mL}}{85.0 \, \text{mL}} \times 100 = 58.8\%
]
3. Mole Fraction of Acetone
The mole fraction of acetone is calculated as:
[
\text{Mole fraction of acetone} = \frac{\text{moles of acetone}}{\text{moles of acetone} + \text{moles of ethyl alcohol}}
]
Step 1: Calculate the moles of acetone
- Molar mass of acetone (C₃H₆O) = 58.08 g/mol
[
\text{moles of acetone} = \frac{\text{mass of acetone}}{\text{molar mass of acetone}} = \frac{27.65 \, \text{g}}{58.08 \, \text{g/mol}} = 0.476 \, \text{mol}
]
Step 2: Calculate the moles of ethyl alcohol
- Molar mass of ethyl alcohol (C₂H₅OH) = 46.07 g/mol
[
\text{moles of ethyl alcohol} = \frac{\text{mass of ethyl alcohol}}{\text{molar mass of ethyl alcohol}} = \frac{39.45 \, \text{g}}{46.07 \, \text{g/mol}} = 0.857 \, \text{mol}
]
Step 3: Calculate the mole fraction of acetone
[
\text{Mole fraction of acetone} = \frac{0.476 \, \text{mol}}{0.476 \, \text{mol} + 0.857 \, \text{mol}} = \frac{0.476}{1.333} = 0.357
]
Final Answers:
- Mass percent of acetone = 41.2%
- Volume percent of ethyl alcohol = 58.8%
- Mole fraction of acetone = 0.357
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