68 231
Assignments Done
99%
Successfully Done
In November 2018

Answer to Question #55278 in Physical Chemistry for manu

Question #55278
ON adding more solid solute to a solution,does solution will freeze more rapidly ?
Please tell me about it , I am very confused it should not more rapidly freeze a/c to me
Expert's answer
Due to colligative properties of solutions there is depression of freezing point of solution in comparison with pure solvent. It means that the freezing point of a pure solvent is lowered by the addition of a solute. Mathematical law for this phenomenon is
ΔTF=KF·b·i (for ideal solutions)
where ΔTF – the freezing point depression,
ΔTF =TF (pure solvent) TF (solution).
KF – the cryoscopic constant, which is dependent on the properties of the solvent.
b – the molality (mol solute per kg of solvent)
i – the van 't Hoff factor (number of ion particles per individual molecule of solute).
For a more accurate calculation at a higher concentration, equation proposed by Ge and Wang should be used. It includes activity of solute which depends on concentration.
So more solid solution – more freezing point depression and the time for crystallization of such solution will be increased.
Answer: solution will not freeze more rapidly because you should cool it to lower temperature

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Assignment Expert
13.10.15, 12:23

Dear kavita, please use panel for submitting new questions

kavita
12.10.15, 19:48

Which of the following substances is good reducing agent?
a. HI
b.KBr
c.FeCl3
d.KClO3
Please tell me its answer. I have a doubt why KBr or KClO3 are not good reducing agent than HI . thank you very much

Assignment Expert
05.10.15, 16:40

Dear manu,
You are welcome. We are glad to be helpful. If you really liked our services please press like button beside answer field. Thank you!

manu
05.10.15, 14:53

Thanks a lot , for detailed explanation ,

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions