🤓 Based on our data, we think this question is relevant for Professor Kulatunga's class at USF.
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.