Solutions | Freezing Point Depression | Chemistry Class 12 | By. Mrs Shubhada Walawalkar

It discusses briefly about Freezing Point Depression,Relation between Freezing Point Depression and Molarity and Molar Mass of Solute and Numerical on Freezing Point Depression.

1. 2.SOLUTIONS
Colligative properties of non electrolyte solutions
3.Freezing point depression
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2. ➢Freezing point.
➢Freezing point depression
➢Freezing point depression as
consequences of vapourpressure.
➢Freezing point depression and
concentration of solute.
➢Molar mass of solute fromfreezing
point depression.
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3. 2.9(1) Freezing point
Freezing point of a liquid is the
temperatureat which liquid and
solid are in equilibriumand the two
phases have the same vapour
pressure.
H
2O(S)
ice
H
2O(l)
water
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4. 0
0
c 100
0
c
Ice Water
Freezing Boiling
Point point
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5. MOLECULES IN
LIQUID STATE
MOLECULES IN
SOLID STATE
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8. 2.9(2) -Depression in
Freezing Point:
The freezing point of a
solvent is lowered by
dissolving a
nonvolatile solute
into it.
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10. If ??????
0
f= freezing point of pure
solvent and,
??????
�= freezing point of solution
in which nonvolatile solute is
dissolved.
T
0
�> ??????
�
∆ ??????
�= T
0
�-??????
�
A
B
C
D
E
F
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11. Freezing point depression as a
consequence of vapourpressure
lowering:
Three curves from diagram:
1. AB= vapourpressure curve of solid
solvent.
2. CD= vapourpressure curve of pure
liquid solvent.
3. EF= vapourpressure curve of
solution
A
B
C
E
F
D
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12. *The curves AB& CDintersect at
point Bwhere solid & liquid
phases of pure solvent are in
equilibrium.
At point Btwo phases with same
vapourpressure.The temperature
corresponding to Bis the freezing
point of solvent(??????
0
f)
A
B
D
C
E
F
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13. The curves EF& ABintersect at point
Ewhere solid solvent and solutionare
in equilibrium.
At point Etwo phases with same
vapourpressure.
The temperature corresponding to Eis
the freezing point of solution (??????�)
??????
&#3627408467;< ??????
0
f
A
B
D
C
E
F
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14. At the freezing point of a pure
liquid the attractive forces
among molecules are large
enough to cause the change of
phase from liquid to solid.
Hence, ??????
&#3627408467;< ??????
0
&#3627408467;
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15. In a solution, separation of solvent
molecules is more than that of
pure solvent. This results in
decreasing the attractive forces
between solvent molecules. The
temperature of the solution is
lowered below the freezing point
of solvent to cause the phase
change.
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16. Freezing point depression and concentration of solute:
For a dilute solution,
The freezing point depression,
∆T
f α m ⸫∆T
f= K
fm -----(1)
K
f= proportionality constant
= freezing point depression constant
= cryoscopicconstant.
If m=1, ∆T
f= K
f
Define K
f cryoscopicconstant as, the depression in freezing
point produced by1 molalsolution of a nonvolatile solute.
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17. Unit of K f :
K f = ∆??????&#3627408467;/ &#3627408474;
= ??????????????????℃/ &#3627408474;??????&#3627408473;/ &#3627408472;&#3627408468;
= K kg mol
-1
or ℃ kg mol -1
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18. Molar mass of solute from freezing point depression:
∆T
f= K
f m -----(1).
Suppose we prepare a solution by dissolving W
2gof solutein
W
1gof solvent.
Moles of solute in W
1gof solvent = ??????
2/??????
2
M
2is the molar mass of solute.
Mass of solvent = W
1g= ??????
1&#3627408468;/1000 &#3627408468;&#3627408472;&#3627408468;
= ??????
1/1000 &#3627408472;&#3627408468;
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19. Molality m = moles of solute/mass of solvent in kg
= ??????
2/??????
2&#3627408474;??????&#3627408473;
(??????
1/ 1000)&#3627408472;&#3627408468;
= 1000 ??????
2
??????
2??????
1mol kg
-1
Substitution of this value in eq. (1) ∆T
f= K
fm
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20. Substitution of this value in eq. (1)
∆T
f= K
fm
∆T
f= K
f×1000 ×W
2
??????
2 ×??????
1
M
2= 1000 ×??????
&#3627408467;×??????
2
∆??????
&#3627408467;×??????
1
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21. Numerical on freezing point depression.
A 5% (by mass)aqueous solution of cane sugar has freezing point
271K.Calculate the freezing point of 5% aqueous glucose
solution.(molar mass of ofcane sugar= 342 g/mol)
Given:
% by mass of cane sugar= 5%
% by mass of glucose = 5%
Freezing pointof cane sugar
solution = 271K.
Molar mass of cane sugar =
342 g/mol
Freezing point of glucose
solution.=?
Solution:
Formula:
ΔT
f=1000 ×K
f×W
2
M
2×W
1
⸫K
f= ΔT
f ×M
2×W
1
1000 ×W
2
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22. ⸫K
f= ΔT
f×M
2×W
1
1000 ×W
2
= 2 ×342 ×95
1000 ×5
=12.99 K Kg /mol
ΔT
f 2 = K
f×W’
2×1000
M’
2×W’
1
= 12.99 ×5 ×1000
180 ×95
Freezing point of cane sugar = 271K
Freezing point of water = 273K
⸫ΔT
f= T
0
&#3627408467;-T
f
= 273K-271
=2 K
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23. =12.99 ×5 ×1000
180 ×95
= 3.801 K
⸫freezing point of solutionT
f 2 =T
0
f-ΔT
f 2
= 273 -3.801
= 269.2K
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24. Thank you
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