posted by Nikita on .
Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at -14.5 degrees celcius ? The freezing point for pure water is 0.0 degrees celcius and Kf is equal to 1.86 degrees celcius/m.
If the 3.90 m solution from Part A boils at 103.45 degrees celcius, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 degrees celcius and Kb is equal to 0.512 degrees celcius/m
Did you work the first part to obtain 3.90m for the KNO3 solution. Then
m = moles/kg solvent.
Solve for moles.
Moles = grams/molar mass
Solve for grams. I think it is about 110 g or so but you need to work it through.
(103.45-100) = i*0.512*3.90
Solve for i.