Question

A chemist combined chloroform (CHCI3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform, Xchloem, is 0.219. The densities of chloroform and acetone are 1.48g/ml, and 0.791 g/ml, respectively. 

Calculate the molarity of the solution.

Calculate the molality of the solution.

Answer 1

Since the mole fraction of chloroform = 0.219, It means,

0.219 = (moles chloroform) / (moles chloroform + moles acetone)

Assume we have exactly 1 mole of chloroform, and let x = moles acetone. Then:
0.219 = 1/(1+x)
1+x = 1/0.219 = 4.566
x = 3.566 moles acetone
Molecular weight of Acetone= 58.1 g/mol
Mass of 3.566 moles acetone = 3.566 mol X 58.1 g/mol = 207.18 g acetone

molality = moles solute / kilogram solvent = 1 mol CHCl3 / 0.20718 kg acetone = 4.827 molal

Molecular weight of CHCL3= 119.4 g/mol
Molarity = moles solute / L of solution
Volume CHCl3 = 1 mol X 119.4 g/mol / 1.48 g/mL = 80.7 mL
Volume acetone = 207.18 g / 0.791 g/mL = 261.92 mL

Assuming that these two liquids mix ideally (the volume of the solution is equal to the sum of their individual volumes), then the total volume is 342.62 mL.

So, the molarity of CHCl3 in the solution is 1 mol / 0.342 L = 2.924 Molar