Calculating Freezing Point of K2CrO4 Solution

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SUMMARY

The freezing point of a solution made by dissolving 250.0 g of potassium chromate (K2CrO4) in 1.00 kg of water is calculated using the formula ΔT(freezing point) = (Van't Hoff Factor)(molal concentration of solute particles)(molal freezing point-depression constant). The correct Van't Hoff Factor for K2CrO4 is 3, leading to a calculated change in freezing point of -5.75 degrees Celsius. However, the discrepancy with the answer key, which states -7.18 degrees Celsius, is resolved by recognizing that the calculation should only consider 1 kg of solvent, not 1.25 kg.

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Homework Statement


A solution is made by dissolving 250.0 g of solid potassium chromate (K2CrO4) in 1.00 kg of water?
what will be the freezing point of the new solution.



Homework 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)

molality= moles of solute/ kg of solution

The Attempt at a Solution


(250 g K2CrO4)(1 mol K2CrO4/194.188 K2CrO4) = 1.29 mol of K2CrO4
(1.29 mol K2CrO4/1.25 kg solution) = 1.0296 mol/kg
Change in freezing point = (1.0296 molality)(1.86 degrees Celsius/molality)(3 Van't Hoff Factor)
Change in freezing point = 0 degrees celsius - 5.750 degrees celsius
Change in freezing point = -5.75 degrees celsius

though the answer keys tells me its -7.18? please help me out
 
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uuuhhhx2 said:

Homework Statement


A solution is made by dissolving 250.0 g of solid potassium chromate (K2CrO4) in 1.00 kg of water?
what will be the freezing point of the new solution.



Homework 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)

molality= moles of solute/ kg of solution

The Attempt at a Solution


(250 g K2CrO4)(1 mol K2CrO4/194.188 K2CrO4) = 1.29 mol of K2CrO4
(1.29 mol K2CrO4/1.25 kg solution) = 1.0296 mol/kg
Change in freezing point = (1.0296 molality)(1.86 degrees Celsius/molality)(3 Van't Hoff Factor)
Change in freezing point = 0 degrees celsius - 5.750 degrees celsius
Change in freezing point = -5.75 degrees celsius

though the answer keys tells me its -7.18? please help me out

1kg of solvent only, not 1.25kg
 
nvm i got it
 

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