Calculating Molality and Mass of KBr for Freezing Point Depression

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SUMMARY

The discussion focuses on calculating the mass of potassium bromide (KBr) needed to lower the freezing point of 535 ml of water by 1.40 degrees Celsius. The freezing point depression formula used is Tf = Kf * m, where Kf for water is 1.86 °C kg/mol. Initially, the user calculated a molality of 0.75, leading to an incorrect mass of 47.6 g. After considering the van 't Hoff factor (i = 2 for KBr), the user recalculated and found the correct mass to be 23.9 g.

PREREQUISITES
  • Understanding of freezing point depression and colligative properties
  • Familiarity with the van 't Hoff factor and its application
  • Basic knowledge of molality and molar mass calculations
  • Ability to perform unit conversions, particularly between grams and kilograms
NEXT STEPS
  • Study the van 't Hoff factor and its significance in colligative properties
  • Learn about freezing point depression calculations in detail
  • Explore the properties of ionic compounds and their dissociation in solution
  • Practice additional problems involving molality and freezing point depression
USEFUL FOR

Chemistry students, educators, and anyone involved in physical chemistry or solution chemistry who seeks to understand colligative properties and their calculations.

pari786
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Assume that the density of water is 1.00 g/ml . Calculate the quantity of KBr that would need to be added to lower the freezing point of 535 ml of water by 1.40 degree celsius:

What I'm trying to do is the following:

Step 1:

Tf = Kf . m

I took Tf = 1.4
Kf = 1.86
and found the molality as 0.75

Step 2:

then I used the molality equation which is

molality = moles solute/kg solvent

molality = 0.75
kg solvent = 0.535 kg
and got the no. of moles as 119 g/mol


Step 3


no. of moles = mass /molar mass and
got 47.6 g for mass

but the answer is still wrong ... I don't know where I'm making mistake. It's an online assignment so I don't know what the answer should be. But please guys help me and check if I'm doing everything right.
 
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pari786 said:
Assume that the density of water is 1.00 g/ml . Calculate the quantity of KBr that would need to be added to lower the freezing point of 535 ml of water by 1.40 degree celsius:

What I'm trying to do is the following:

Step 1:

Tf = Kf . m

I took Tf = 1.4
Kf = 1.86
and found the molality as 0.75

Step 2:

then I used the molality equation which is

molality = moles solute/kg solvent

molality = 0.75
kg solvent = 0.535 kg
and got the no. of moles as 119 g/mol


Step 3


no. of moles = mass /molar mass and
got 47.6 g for mass

but the answer is still wrong ... I don't know where I'm making mistake. It's an online assignment so I don't know what the answer should be. But please guys help me and check if I'm doing everything right.


what is the vanthoff factor for KBr?
 
gravenewworld said:
what is the vanthoff factor for KBr?

oh so you mean to say that I do have to consider vanthoff factor for KBr too in this problem?
I must have forgotten about that?
 
pari786 said:
oh so you mean to say that I do have to consider vanthoff factor for KBr too in this problem?
I must have forgotten about that?

vanthoff factor for KBr is 2 since we get two ions from one mole of KBr.
Is that right?
 
pari786 said:
vanthoff factor for KBr is 2 since we get two ions from one mole of KBr.
Is that right?

Hi, I'm doing the caculations again, see if they are right now,
I'm getting molality as 0.376
no. of moles =0.20116
molar mass =119 g/mol
mass = 23.9 g
 
hey, thanks for reminding me about the i value ... I tried it in the assignment questions and i got the right answer.
Thanks a lot for the help.
 

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