Asked by Brunette
Compound: Concentration: Measured deltaT
LiCl 6.00 g/kg 0.492 deg C
HCl 4.00 g/kg 0.389 deg C
NaCl 6.00 g/kg 0.356 deg C
Determine the value of the vant hoff factor for each salt for the experimentally measured delta T value. (kf=1.86)
I have tried to figure this out, but it is saying I'm wrong. I might be doing the math wrong. Can you help me?
LiCl 6.00 g/kg 0.492 deg C
HCl 4.00 g/kg 0.389 deg C
NaCl 6.00 g/kg 0.356 deg C
Determine the value of the vant hoff factor for each salt for the experimentally measured delta T value. (kf=1.86)
I have tried to figure this out, but it is saying I'm wrong. I might be doing the math wrong. Can you help me?
Answers
Answered by
DrBob222
dT = i*Kf*m
i = dT/(Kf*m)
dT = 0.492 for LiCl
Kf = 1.86 (I assume you are using this value or one given in class to use.
m = mols/kg solvent
mols = 6.00/molar mass LiCl
I get 1.869 for i which would round to 1.87.
If you post your work for one of the problems I will find your error for you.
i = dT/(Kf*m)
dT = 0.492 for LiCl
Kf = 1.86 (I assume you are using this value or one given in class to use.
m = mols/kg solvent
mols = 6.00/molar mass LiCl
I get 1.869 for i which would round to 1.87.
If you post your work for one of the problems I will find your error for you.
Answered by
Meeeee
I calculate the amount of kg, which is .00600 kg. Then I calculate the molality:
.14 mols/.00600kg = 23.3 m
23.3*1.86=43.338
.492/43.338=.0114
I don't know which step is incorrect.
.14 mols/.00600kg = 23.3 m
23.3*1.86=43.338
.492/43.338=.0114
I don't know which step is incorrect.
Answered by
DrBob222
The way I read the LiCl part of the problem, (LiCl) = 6.00 g/kg ; therefore, mols LiCl = 6.00 g/42.394 = 0.1415
m = 0.1415 mols/1 kg solvent.
i = 0.492/(0.1415*1.86) = 1.869 which I rounded to 1.87. Therefore, i isn't quite 2 as we would predict for two particles (Li^+ and Cl^-) but just less than that. Your error is in calling the solvent 0.006 kg; the problem says it is 1 kg (at least 6.00 g LiCl/kg says that to me). That means to me you would have 12.00 g/2 kg, 18.00 g/3 kg etc.
m = 0.1415 mols/1 kg solvent.
i = 0.492/(0.1415*1.86) = 1.869 which I rounded to 1.87. Therefore, i isn't quite 2 as we would predict for two particles (Li^+ and Cl^-) but just less than that. Your error is in calling the solvent 0.006 kg; the problem says it is 1 kg (at least 6.00 g LiCl/kg says that to me). That means to me you would have 12.00 g/2 kg, 18.00 g/3 kg etc.
Answered by
Meeeee
thank you! I figured that was the problem.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.