When 19.5g of FCH2COOH is dissolved in 500g of water. Calculate the depression in freezing point. Kf for water= 1.86 K Kg mol-1
Answers
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Answer:
Fluoroacetic acid has a molecular mass of 78 g/mol.
Number of moles of fluoroacetic acid is
78
19.5
=0.25.
Molality is the number of moles of solute in 1 kg of solvent.
Molality =
1000
500
0.25
=0.50 m
Calculated depression in the freezing point.
ΔT
f
=K
f
×m=1.86×0.50=0.93 K
Van't Hoff factor is the ratio of observed freezing point depression to calculated freezing point depression.
i=
0.93
1.0
=1.0753
Let c be the initial concentration of fluoro acetic acid and α be its degree of dissociation.
c(1−α)
CH
2
FCOOH
→
cα
CH
3
FCOO
−
+
cα
H
+
Total number of moles =c(1−α)+cα+cα=c(1+α)
i=
c
c(1+α)
=1+α=1.0753
α=0.0753
[CH
2
FCOO
−
]=[H
+
]=cα=0.50×0.0753=0.03765
[CH
2
FCOOH]=c(1−α)=0.50(1−0.0753)=0.462
K
a
=
[CH
2
FCOOH]
[CH
2
FCOO
−
][H
+
]
K
a
=
0.462
0.03765×0.03765
K
a
=3.07×10
−3
Hence, the van't Hoff factor is 1.0753 and dissociation constant is 3.07×10
−3
for fluoroacetic acid.
Explanation: