Chemistry, colligative properties-freezing point

[Flip]

Homework Statement

A solution is made by dissolving 250.0 g of solid potassium chromate in 1.00 kg of water. What will be the freezing point of the new solution?

molal freezing point-depression constant of water = 1.86 degrees Celsius/molality
molality = mol/kg

Relevant equations

[Delta]T(freezing point)=(Van't Hoff Factor)(molal concentration of solute particles)(molal freezing point-depression constant)

Van't Hoff Factor = (moles of particles in solution/moles of solute dissolved)The attempt at a solution

(250 g K2CrO4)(1 mol K2CrO4/194.188 K2CrO4) = 1.29 mol of K2CrO4
(1.29 mol K2CrO4/1.00 kg H2O) = 1.29 mol/kg
Change in freezing point = (1.29 molality)(1.86 degrees Celsius/molality)
Change in freezing point = 0 degrees celsius - 2.40 degrees celsius
Change in freezing point = -2.40 degrees celsiusComments

Answer is -7.18 degrees celsius
Since ionic compounds rarely dissociate completely the Van't Hoff Factor has to be used. Except I don't know how to use it...

 

[Borek]

Since ionic compounds rarely dissociate completely

This is false statement. Simple ionic salts almost always dissociate 100%.

the Van't Hoff Factor has to be used. Except I don't know how to use it...

What is van't Hoff factor for this salt (assuming 100% dissociation)?

 

[Flip]

The Van't Hoff Factor isn't given, which is why I thought that I had to find it myself. But if it does dissociate completely then doesn't that mean that the Van't Hoff Factor would just be 1?

 

[Borek]

No. Write equation of dissociation reaction and use the definition you have already posted.