Question

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.

Answer 1

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Answer 2

Since the mole fraction of chloroform = 0.171, that means that:

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

Assume we have exactly 1 mole of chloroform, and let x = moles acetone. Then:
0.171 = 1/(1+x)
1+x = 1/0.171 = 5.85
x = 4.85 moles acetone

Mass of 4.85 moles acetone = 4.85 mol X 58.1 g/mol = 282 g acetone

molality = moles solute / kilogram solvent = 1 mol CHCl3 / 0.282 kg acetone = 3.55 molal

Molarity = moles solute / L of solution
Volume CHCl3 = 1 mol X 119.4 g/mol / 1.48 g/mL = 80.7 mL
Volume acetone = 282 g / 0.791 g/mL = 356.5 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 437 mL.

So, the molarity of CHCl3 in the solution is 1 mol / 0.437 L = 2.29 Molar