An aqueous solution of ethanol is 14.1M C2H5OH. The density of the solution is 0.853g/mL. What is molality of ethanol in the solution? What is mass % of ethanol in solution?
I got the molality which is -21.7 ethanol, but I can't figure out how to get the mass % from the molality?
molality = mols EtOH/kg solvent
Convert mols EtOH to grams. g = mols x molar mass.
Add g EtOH to 1000 for mass solution.
Then mass % = (grams EtOH/total grams)*100 = ?
By the way, I wonder how you obtained -21.7m. A negative number for concentration? I don't think so.
To determine the molality of ethanol in the solution, we need the mass of ethanol and the mass of the solvent (water in this case). The molality (m) is defined as the number of moles of solute per kilogram of solvent.
Given:
- Concentration (C) of ethanol in the solution = 14.1 M
- Density (d) of the solution = 0.853 g/mL
First, let's convert the density of the solution to kg/L:
Density (d) = 0.853 g/mL = 853 g/L = 853 g / 1000 g/kg = 0.853 kg/L
Next, we'll calculate the mass of ethanol by using the concentration and the density:
Mass of ethanol = Concentration * Volume of solution
But we don't know the volume of the solution. To find it, we can assume that the volume of the solution is 1 L, as the concentration is given in moles per liter (M).
Therefore, the mass of ethanol = 14.1 mol/L * 1 L * 46.07 g/mol (molar mass of ethanol)
= 644.34 g
Now, to calculate the molality of ethanol (m), we need the mass of the solvent (water). Assuming the solution is composed only of ethanol and water:
Mass of the solvent = Total mass of the solution - Mass of ethanol
Since we don't have the total mass of the solution, we can use the density to calculate it:
Total mass of the solution = Volume of solution * Density
Assuming a volume of 1 L, the total mass of the solution = 1 L * 0.853 kg/L = 0.853 kg
Mass of the solvent = 0.853 kg - 0.64434 kg = 0.20866 kg
Finally, we can calculate the molality of ethanol (m):
Molality (m) = Moles of solute / Mass of the solvent
Moles of solute = Concentration * Volume of solution = 14.1 mol/L * 1 L = 14.1 mol
Molality (m) = 14.1 mol / 0.20866 kg = 67.57 mol/kg
Now, to calculate the mass percent of ethanol in the solution, we can use the equation:
Mass percent = (Mass of solute / Mass of solution) * 100
Given the mass of ethanol as 644.34 g (obtained before) and the total mass of the solution as 0.853 kg (also obtained before):
Mass percent of ethanol = (644.34 g / 0.853 kg) * 100 = 75.53%
Therefore, the molality of ethanol in the solution is approximately 67.57 mol/kg, and the mass percent of ethanol in the solution is approximately 75.53%.