Question

Your lab partner combined chloroform, CHCl3, and acetone, (CH3)2CO, to create a solution where the mole fraction of chloroform, Xchloroform, is 0.155. The densities of chloroform and acetone are 1.48 g/mL and 0.791 g/mL, respectively. *** calculate the molarity and molality of the solution *** question has been edited to provide additional information

Answer 1

the Xchloroform being so low indicates that the solvent is probably acetone

that being said start by calculating the moles of chloroform and acetone

Xchloroform = 0.155 = (moles of chloroform)/(moles of chloroform + moles of acetone)

for the purposes of this calculation you can assume moles of chloroform = 1 giving us

0.155 = (1)/(1 + x)

Answer 2

once you have moles of acetone, convert moles of acetone to kg using the molar mass ( be careful, molar masses are in g/mol and you want kg)

molality = 1 mol chloroform/whatever kg you got for acetone

Answer 3

molarity gets a bit trickier, since we want

molality = moles of chloroform/L of solution

convert moles of chloroform + moles of acetone to volume of solution first, add the volumes, and then calculate molality