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
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)
Mass = 1 kg ethanol
Density of ethanol = 0.789 g/cm3
Calculate the volume of ethanol:
▪ convert 1 kg = 103 g
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.
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Mole Fraction concept. You can view video lessons to learn Mole Fraction. Or if you need more Mole Fraction practice, you can also practice Mole Fraction practice problems.
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Based on our data, we think this problem is relevant for Professor Kulatunga's class at USF.
What textbook is this problem found in?
Our data indicates that this problem or a close variation was asked in Chemistry - Zumdahl 8th Edition. You can also practice Chemistry - Zumdahl 8th Edition practice problems.