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AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper
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© examsnet.com
Question : 141
Total: 160
12.25 g of
C
H
3
C
H
2
CHClCOOH is added to 250 g of water to make a solution. If the dissociation constant of above acid is 1.44 ×
10
−
3
,
the depression in freezing point of water in °C is
(
K
f
for water is 1.86 K kg
m
o
l
−
1
)
0.789
0.394
1.183
0.592
Validate
Solution:
First, we will calculate molarity as follows:
Molarity
(
C
)
=
Mass
of
solute
Molar
mass
of
solute
×
Mass
of
solvent
×
1000
=
12.25
×
1000
122.5
×
250
=
0.40
Now, the degree of ionization is calculated as:
α
=
√
K
a
C
=
√
1.44
×
10
−
3
0.4
=
0.66
Now, Van’t Hoff factor (i) is calculated as:
C
H
3
C
H
2
C
H
C
l
C
O
O
H
⇌
C
H
3
C
H
2
C
H
C
l
C
O
O
−
+
H
+
Initial
conc
1
0
0
Equilibrium
1
−
α
α
α
Total number of moles is equal to
1
−
α
+
α
+
α
=
1
+
α
i
=
Moles
after
dissociation
Moles
before
dissociation
=
1
+
α
1
=
1
+
α
Since, value of α is 0.06
Therefore,
i
=
1
+
0.06
=
1.06
The change in freezing point is calculated as follows:
Δ
T
f
=
i
K
f
m
=
1.06
×
0.4
×
1.86
=
0.789
°
C
© examsnet.com
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