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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...
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...
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