Solutions | Boiling point elevation | Chemistry Class 12 | By. Mrs. Shubhada Walawalkar

This video discusses about colligative property- boiling point elevation and numericals based on it.

1. 2. SOLUTIONS
Part -3
Colligative properties of nonelectrolyte
solutions.
2.8 Boiling point elevation
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Colligative
properties
Boiling point Elevation
Freezing point
depression
Osmotic pressure
Vapour
pressure
lowering

3. 2.Boiling Point Elevation
Boiling point(B.P.):-
The boiling point of a liquid is that temperature at which its
vapourpressure becomes equal to the applied pressure.
For liquids in open container applied pressure is atmospheric
pressurei.e.760 mm of Hg. or101.3kPa or1 atm.
It is characteristic of particular liquid
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5. B.P. depends on
1)Pressure
.
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6. B.P. depends on
2)Intermolecular forces
➢At the boiling point of liquid,
its v.p.is equal to 1 atm.
➢If liquid has high v.p.it has low
b.p.(weak intermolecular
forces)
➢If liquid has low v.p.it has high
b.p.(strong intermolecular
forces)
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7. Boiling point elevation:-
The boiling point of a solvent
is elevated by dissolving a
nonvolatile soluteinto it.
i.e. such solution boil at
temperature higher than the
b.p. of pure solvent.
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8. T
0
b= B.P. of pure solvent
T
b= B.P. of a solution
T
b> T
0
b
∆T = T
b -T
0
b -----(1)
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A
B
A
D
C

9. Boiling point elevation as consequence of
vapourpressure lowering:
Comparison of v.p.of pure solvent and
solution containing nonvolatile solute.
At any temperature the vapourpressure of
solution is lower than that of pure solvent.
Hence, the vapourpressure-temperature
curve of solution (CD) lies below that of
solvent (AB).
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A
B
C
D

10. The difference in V.P. ∆P = x
2P
0
1and
Figure.
Indicate that the boiling point (T
b) of
the solution is higher than that of pure
solvent (T
0
b)
At the boiling point of a liquid its
vapourpressure is equal to 1atm.
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11. Boiling point elevation and concentration of solute:
∆T
bα m or ∆T
b= K
bm -----(1)
m = molality of solution,
Kb= proportionality constant called boiling point elevation
constantor molalelevation constant or ebullioscopic
constant.
If m=1, ∆T
b= K
b
Ebullioscopicconstant is the boiling point elevation
produced by 1 molalsolution.
.
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12. Units of Kb:
K
b= ∆T
b/ �
= ??????
mol/kg
= K kg mol
-1
.
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13. Molar mass of solute from boiling point elevation:
∆T
b= K
bm
Suppose we prepare a solution by dissolving W
2g of solute in
W
1g of solvent
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Moles of solutein W
1g of solvent = ??????
2/ ??????
2
M
2is the molar mass of solute.
Mass of solvent = W
1g
= ??????
1??????/ 1000 ??????/�??????
= ??????
1/1000 �??????

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Molality( m)
= moles of solute/mass of solvent in kg
= ??????
2/ ??????
2�??????�
(??????
1/ 1000)�??????
= 1000??????
2
??????
2??????
1mol kg-1

15. Substitution of this value in eq. (1) ∆T
b= K
bmgives,
∆T
b= K
b1000 ??????
2
??????
2??????
1
M
2= 1000??????
????????????
2
∆??????
????????????
1
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16. Numerical on boiling point elevation(11.page no.46 from text book)
A solution of citric acid C
6H
8O
7in 50gram of acetic acid has
boiling point elevation of 1.76KIf Kbfor acetic acid is 3.07K
kg/mol. What is the molality of solution?
Given:-
W
1=50 g acetic acid (solvent)
Δ T
b=1.76K
K
b= 3.07K kg/mol.
m =?
Solution:-
Δ T
b= K
b×m ⸫m= Δ T
b/ K
b
= 1.76/3.07
= ?
= 0.5733 m
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17. Thank You
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