Your lab partner combined chloroform (CHCls) and acetone (C3HO) to create a solu
ID: 1028515 • Letter: Y
Question
Your lab partner combined chloroform (CHCls) and acetone (C3HO) to create a solution where the moleMapdto fraction of chloroform, Xchlaroform, is 0.171. 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 IIn Previous Give Up & View Solution e) Check Answer Next Exit Hin Start this problem by finding the moles of solute and solvent. The question did not specify an amount, mass, or volume of solute or solvent, but the mole fraction of chloroform is given. The mole fraction, X, is the ratio of the number of moles of substance (solute or solvent) in solution to the total number of moles (solute plus solvent) in solution For any value chosen for the total number of moles in solution, the number of moles of substance will be present in a ratio defined by the mole fraction. Remind yourself of the definitions of the molarity and molality.Explanation / Answer
From the question we haveThe mole fraction of chloroform = 0.171,
Therefore
0.171 = (moles chloroform) / (moles chloroform + moles acetone)
Assuming that we have exactly 1 mole of chloroform, and letus assume x = moles acetone.
Now
0.171 = 1/(1+x)
1+x = 1/0.171 = 5.85
x = 4.85 moles acetone
4.85 moles acetone mass = 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 sum of their individual volumes is equal to the volume of the solution), then the total volume is 437 mL.
Hence, the molarity of CHCl3 in the solution is 1 mol / 0.437 L = 2.29 Molar
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.