Colligative properties are those that depend on the number of solute particles.
ID: 979959 • Letter: C
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,
Tf = Kfmi
Tb = Kbmi
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:
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 !!!
108 gExplanation / Answer
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