A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform

Question

A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform, 𝜒chloroform, is 0.171 . The densities of chloroform and acetone are 1.48 g/mL and 0.791 g/mL, respectively. Calculate the molarity of the solution. Assume the volumes are additive. Then calculate the molality of the solution.

Answer 1

Let's begin with our definitions of molarity and molality:

Molarity = moles of solute / liter of solution

Molality = moles of solute / kg of solvent

We have two liquids (chloroform and acetone), so which is the solvent and which is the solute? In a case like this, we generally take the one that is present in the greatest amount and make that the solvent. In this case that would be acetone, since its mole fraction (and volume) is greater than than of chloroform.

X_chloroform = 0.171

X_acetone = 1 - 0.171 = 0.829

To calculate molarity and molality we will need to know the MOLES of each ingredient. We will also need to know the final volume (for molarity) and the mass of solvent (acetone) for molality.

We can start by assuming we have a total of 1 mole. Thus we have 0.171 moles chloroform and 0.829 moles acetone. From this, we can find the mass of each and the volume of each (using densities)

mass chloroform (CHCl₃) = 0.171 mols x 119.37 g / mol = 20.41 g CHCl₃

mass acetone (C₃H₆O) = 0.829 mols x 58.08 g / mole = 48.15 g C₃H₆O

volume CHCl₃ = 20.41 g x 1 ml / 1.48 g = 13.79 mls (solute)

volume C₃H₆O = 48.15 g x 1 ml / 0.791 = 60.87 mls (solvent)

Molarity = moles of CHCl₃ / liters of solution:

moles CHCl₃ = 0.171 mols

liters of solution = 13.79 mls + 60.87 mls = 74.66 mls x 1 L / 1000 ml = 0.07466 L

Molarity = 0.171 mols / 0.07466 L = 2.29 M

Molality = moles of CHCl3 / kg solvent

moles CHCl3 = 0.171 mols

kg of solvent (acetone) = 48.15 g x 1 kg / 1000 g = 0.04815 kg

Molality = 0.171 mols / 0.04825 kg = 3.55 m