Question

Colligative properties are those that depend on the number of solute particles. Because electrolytes dissociate into ions, the concentration of particles in the solution is greater than the formula-unit concentration of the solution. For example, if 1 mol of Na2SO4 totally dissociates, 3 mol of ions are produced (2 mol of Na+ ions and 1 molof SO42− ions). Thus, a colligative property such as osmotic pressure will be three times greater for a 1 M Na2SO4 solution than for a 1 Mnonelectrolyte solution.

However, complete dissociation of electrolytes does not always occur. The extent of dissociation is expressed by the van't Hoff factor, i:

i = moles of particles in solutionmoles of solute dissolved

The equations for colligative properties can be written to include i. For example,

ΔTf = Kf⋅m⋅i

ΔTb = Kb⋅m⋅i

Part A

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mLof water to produce a solution that freezes at −14.5 ∘C? The freezing point for pure water is 0.0 ∘Cand Kf is equal to 1.86 ∘C/m.

Express your answer to three significant figures and include the appropriate units.

PART A ANSWER:

108 g

Part B

If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 ∘C and Kb is equal to 0.512 ∘C/m.

Express your answer numerically.

HELP WITH PART B !!!

Answer 1

Molarity of solution of KNO3 = 3.90 m

Mass of KNO3 =108 g

Boiling point = 103.45 deg C . boiling point of pure water = 100.00 deg C

Kb of water = 0.512 deg C /m

Delta Tb = i m kb

Delta Tb = elevation in boiling point. i is vant’ hoff factor. Kb is boiling point constant of water solvent.

Elevation in boiling point = boiling point of the solution – boiling point of the solvent

Formula:

= 103.45 deg C – 100.00 deg C

= 3.45 deg C

3.45 deg C = i    x 3.90 m x 0.512 deg C / m

i = 1.73