Chemistry Stochiometry

Question

A chemist combined chloroform (CHCl3)(CHCl3) and acetone (C3H6O)(C3⁢H6O) to create a solution where the mole fraction of chloroform, 𝜒chloroform,�chloroform, is 0.2190.219. 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.

Calculate molality and molarity.

Answer 1

mole fraction chloroform = 0.219

mole fraction acetone = 1.000 - 0.219 = 0.781

Let's find the volume of solution:

0.219 mols chloroform x 119.4 g / mol = 26.15 g chloroform x 1 ml / 1.48 g = 17.7 mls chloroform

0.781 mols acetone x 58.08 g / mol = 45.36 g acetone x 1 ml / 0.791 g = 57.3 mls acetone

total volume = 17.7 mls + 57.3 mls = 75 mls of solution (0.075 liters)

Next, we will find the molarity (M) of each component:

molarity = moles / liter

molarity of chloroform = 0.219 mols / 0.075 L = 2.92 M

molarity of acetone = 0.781 mols / 0.075 L = 10.4 M

Finally, we will find the molality (m):

molality (m) = moles solute / kg solvent

we have 2 liquids so we will choose the one in largest amount as the solvent. That would be acetone.

kg of solvent (acetone) = 45.36 g acetone x 1 kg / 1000 g = 0.04536 kg

moles of solute (chloroform) = 0.219 mols

molality of chloroform = 0.219 mols / 0.0454 kg = 4.82 m