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.
