Question

Your lab partner combined chloroform and acetone to create a solution where the mole fraction of chloroform, Xchloroform, is 0.171. The densities of chloroform and acetone are 1.48 g/mL and 0.791 g/mL, respectively.1. Calculate the molality of the solution.2. Calculate the molarity of the solution.

Answer 1

Answer:

Explanation:

[u]Assumptions[/u]

1. There is exactly 1 mole of chloroform

2. The liquids mix together well such that the volume of the solution is the sum of the volumes of the two liquids

Given that the mole fraction of the Chloroform is 0.171

Mole fraction of Chloroform =

Mole of chloroform /(Mole of chloroform + mole of acetone)

According to assumptions, mole of chloroform is equal to 1

Therefore 0.171 =1/(1+mole of acetone)

1 + mole of acetone = 1/0.171

Mole of acetone = 1(/0.171) - 1

Moles of acetone = 4.85mol.

From Stochiometry

Mass of acetone = Mole of acetone * Molar Mass of acetone

Molar mass of acetone = 58.1grams/mol

Mass of acetone = 4.85 *58.1 = 282g =0.282kg

Mass of chloroform = moles of chloroform *Molar mass of Chloroform

Molar mass of chloform = 119.4 grams/mol

Mass of chloroform = 1* 119.4 =119.4g=0.1994kg

Volume of acetone = Mass of acetone / Density of acetone

Volume of acetone = 282/0.791

Volume of acetone = 357mL

Volume of Chloroform = Mass of Chloroform /Density of Chloroform

Volume of Chloroform = 119.4/1.48

= 81mL

Total volume of solution = 357mL+81mL = 438mL = 0.438L

1. Molarity = Moles of solute(chloroform)/mass of solvent (acetone) in kg

Molarity = 1/0.282 = 3.55molal

2. Molarity = moles of solute( chloroform) /Volume of solution

= 1/0.438 =2.28Molar

Therefore the molality and molarity respectively are 3.55 and 2.28.