chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform, 𝜒chloroform, is 0.203 .

Question

The densities of chloroform and acetone are 1.48 g/mL1.48 g/mL and 0.791 g/mL,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

0.203 mols CHCl3

1 - 0.203 = 0.797 mols C3H6O

mass of CHCl₃ = 0.203 mols x 119.4 g / mol = 24.24 g

volume of CHCl₃ = 24.24 g x 1 ml / 1.48 g = 16.38 ml

mass C₃H₆O = 0.797 mols x 58.1 g /mol = 46.31 g

volume of C₃H₆O = 46.31 g x 1 ml / 0.791 g = 58.55 ml

Total volume = 16.38 ml + 58.55 ml = 74.93 ml = 0.07493 L

Total kilograms = 24.24 g + 46.31 g = 70.55 g = 0.07055 kg

Molarity = moles / liter

CHCl₃ = 0.203 mols / 0.07493 L = 2.71 M CHCl₃

C₃H₆O = 0.797 mols / 0.07493 L = 10.6 M C₃H₆O

Molality = mols / kg

CHCl₃ = 0.203 mols / 0.07055 kg = 2.88 m CHCl₃

C₃H₆O = 0.797 mols / 0.07055 kg = 11.3 m C₃H₆O