2.44 gram of Benzoic acid
dissolved in benzine to form
Solution ? then find out
molarity
and formality ?
500ml
Answers
Answered by
0
Answer:
Correct option is
A
99.2 %
Given,
W
B
=2g,K
f
=4.9KKgmol
−1
W
A
=25g,ΔT
f
=1.62K
Now,
ΔT
f
=K
f
×
M
B
W
B
×
W
A
1000
M
B
=
1.62×25
4.9×2×1000
=241.98g/mol
2C
6
H
5
COOH⇌(C
6
H
5
COOH)
2
If x is the degree of association, (1−x) mole of benzoic acid left undissociated & corresponding x/2 as associated moles of C
6
H
5
COOH at equilibrium.
∴ Total number of moles of particles at eqm,
⇒1−x+
2
x
=1−
2
x
=i
⇒i=
Abnormalmolecularmass
Normalmolecularmass
⇒1−
2
x
=
241.98
122
⇒
2
x
=1−
241.98
122
⇒x=0.992=99.2%
∴ Degree of association of benzene =99.2%
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