A solution of acetic acid, CH3COOH, has a concentration of 0.218 M and a density of 1.00 g mL-1. What is the molality of this solution?
To find the molality (m) of a solution, we need to determine the moles of solute (acetic acid) per kilogram of solvent (water).
First, we calculate the moles of acetic acid using its concentration (C) and volume (V):
Moles of acetic acid (n) = concentration (C) * volume (V)
Given that the acetic acid solution has a concentration of 0.218 M, we can assume a volume of 1L (since 1L of solution is approximately equal to 1kg of solution):
n = 0.218 mol/L * 1 L = 0.218 mol
Since the density of the solution is 1.00 g/mL, we can calculate the mass of the solution:
Mass of solution = density * volume
Mass of solution = 1.00 g/mL * 1000 mL = 1000 g
To calculate the mass of the solvent (water), we subtract the mass of the solute (acetic acid) from the mass of the solution:
Mass of solvent = mass of solution - mass of solute
Mass of solvent = 1000 g - 0.218 mol * 60.05 g/mol (molar mass of acetic acid)
Mass of solvent = 1000 g - 13.0699 g = 986.9301 g
Finally, we can calculate the molality (m) using the moles of acetic acid and the mass of the solvent:
Molality (m) = moles of solute / mass of solvent (in kg)
Molality (m) = 0.218 mol / 0.9869301 kg = 0.221 mol/kg
Therefore, the molality of the acetic acid solution is 0.221 mol/kg.
To determine the molality of a solution, we need to know the moles of solute (acetic acid) and the mass of the solvent (in this case, the solution).
First, we need to calculate the moles of acetic acid using the concentration and volume of the solution. However, we cannot directly calculate the volume of the solution from its density. We need additional information such as the mass of a given volume or the total volume of the solution.
If we have either the mass of a given volume or the total volume of the solution, we can calculate the moles of acetic acid and then proceed to find the molality.
Please provide additional information such as the mass of a given volume or the total volume of the solution to proceed with the calculation.
0.22
0.218M means 0.218 mols CH3COOH/L solution.
g CH3COOH = 0.218 x molar mass = approx 13g but you should redo this and all of the other calculations that follow. I've estimated everything.
From density, mass = volume x density = 1000 mL x 1 g/mL = 1000 grams.
So 1000g = mass H2O + mass CH3COOH = mass H2O + 13 g or mass H2O = approx 1000-13 = about 987g.
m = molality = mols/kg solvent
m = 0.218mols/0.987 kg = ?