Question

A chemist combined chloroform ( CHCl 3 ) and acetone ( C 3 H 6 O ) to create a solution where the mole fraction of chloroform, χ chloroform , is 0.203 . The densities of chloroform and acetone are 1.48 g/mL and 0.791 g/mL , respectively. Calculate the molarity of the solution.

Answer 1

Answer:

Molarity is 13,4M

Explanation:

As mole fraction of chloroform is 0,203; mole fraction of acetone will be 1-0,203=0,797

That means that per 100 moles you have 20,3 moles of chloroform and 79,7 moles of acetone.

Using molar mass and density it is possible to know the volume these moles occupy, thus:

Chloroform: 20,3 moles×$\frac{119,38g}{1mol}$×$\frac{1mL}{1,48g}$= 1637mL= 1,637L

Acetone: 79,7 moles×$\frac{58,08g}{1mol}$×$\frac{1mL}{0,791g}$=  5852mL = 5,852L

That means that total volume is 1,637L + 5,852L = 7,489L

As moles are 100, molarity is:

100mol / 7,489L = 13,4M

I hope it helps!