Grade 12Physical Chemistry
Depression in freezing point of 1M urea, 1M NaCl, 1M CaCl2 are in the ratio of
4 Answers
User
Since we know that ΔT = i Kf mThus ΔT is directly related to i ( Vant Hoff Factor) Now, i (urea) = 1i ( NaCl) = 2I (CaCl2) = 3Depression in freezing points are in the ratio of 1:2:3 .
Krish Gupta
Depression in freezing point = i * Kf * m
NaCl has i = 2
In the above formula 'Kf*m' value is same for both Urea , CaCl2 and NaCl.
In the above formula 'Kf*m' value is same for both Urea , CaCl2 and NaCl.
Thus, the only difference that we see is due to ‘i’, the Van’t Hoff factor.
Now, i (urea) = 1i ( NaCl) = 2I (CaCl2) = 3Depression in freezing points are in the ratio of 1:2:3 .
Your question seems to be incomplete but what I would like to let you know is that depression of freezing point is a colligative property is that it is only dependent on the ratio of the number of solute particles present to the number of solvent molecules and not in any way the nature of the chemical species present.
So when handling this question put that highlight in the back of your mind.
Other colligative properties you may also come across are boiling point elevation, those dealing with pressure and osmotic pressure.
Kushagra Madhukar
Dear student,
Please find the solution to your problem below.
Depression in freezing point = i * Kf * m
Now, for the given solutes m is roughly the same and Kf being dependent on solvent is also same, but the van’t Hoff factor is different.
i (urea) = 1; i (NaCl) = 2; i (CaCl2) = 3
Hence ratio of depression in freezing point = 1 : 2 : 3
Hope it helps.
Thanks and regards,
Kushagra
Samera Abdull
Depression in freezing point = i * Kf * m
Now, for the given solutes m is roughly the same and Kf being dependent on solvent is also same, but the van’t Hoff factor is different.
i (urea) = 1; i (NaCl) = 2; i (CaCl2) = 3
Hence ratio of depression in freezing point = 1 : 2 : 3
By s Abdul Rahman
Grade 12
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