Answer to Question #275737 in General Chemistry for Myin

Question

The density at 20 0

C of a 0.500 M solution of acetic acid in water is

1.0042 g/mL. What is the molality of the solution? The molar mass of

acetic acid, CH3CO2H, is 60.05 g/mol.

Answer 1

Let the volume of the acetic acid solution be 1 litre = 1000 ml

thus, moles of acetic acid = molarity*volume of solution in litres = 0.5*1 = 0.5 moles

now, molar mass of acetic acid = 60.05 g/mole

thus, mass of 0.5 mole acetic acid = 30.025 g........(1)

now, total mass of solution = density of solution *volume = 1.0042*1000 = 1004.2 g..........(1)

thus, mass of solvent in the solution = mass of solution - mass of acetic acid = 1004.2 - 30.025 = 974.175 g

= 0.974175 kg

Now, molality = moles of solute/mass of solvent in kg = 0.5/0.974175 = 0.51325M