Question

Your lab partner combined chloroform (CHCl3) and acetone (C3H60) to create a solution where the mole fraction of chloroform, Xchloroform, is 0.219. The densities of chloroform and acetone are 1.48 g/mL and 0.791 g/mL, respectively. Calculate the molarity of the solution. Number Calculate the molality of the solution. Number Imi ITI

Answer 1

Solution-
From the question we have the mole fraction of chloroform = 0.171

Therefoew 0.171 = (moles chloroform) / (moles chloroform + moles acetone)

Let us assume we have exactly 1 mole of chloroform, also assume we have let x = moles acetone. Then:
0.219 = 1/(1+x)
1+x = 1/0.219 =4.57
=>x = 3.57 moles acetone

So the mass of 3.57 moles acetone =3.57 mol* 58.1 g/mol

= 207.417 g acetone

Now the molality = moles solute / kilogram solvent

= 1 mol CHCl3 / 0.207 kg acetone

= 4.83molal

Now the Molarity = moles solute / L of solution
And the volume CHCl3 = 1 mol * 119.4 g/mol / 1.48 g/mL
= 80.7 mL
THe volume acetone = 207.417 g / 0.791 g/mL = 262.22 mL

So assuming that both liquids mix ideally this means that the volume of the solution is equal to the sum of their individual volumes

then the total volume is 342.92 mL.

Hence, the molarity of CHCl3 in the solution is 1 mol / 0.34292 L = 2.92 Molar