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AP EAMCET Engineering 2011 Solved Paper
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© examsnet.com
Question : 53
Total: 160
If the values of
Λ
∞
of
N
H
4
C
l
,
N
a
O
H
and
N
a
C
l
are 130,217 and
109
o
h
m
−
1
c
m
2
equiv
−
1
respectively, the
Λ
∞
of
N
H
4
O
H
in
o
h
m
−
1
c
m
2
equiv
−
1
is
238
196
22
456
Validate
Solution:
Λ
∞
N
H
4
C
l
=
Λ
∞
N
H
4
+
+
Λ
∞
C
l
−
=
130
.......(i)
Λ
∞
N
a
O
H
=
Λ
∞
N
a
+
+
Λ
∞
O
H
−
=
217
........(ii)
Λ
∞
N
a
C
l
=
Λ
∞
N
a
+
+
Λ
∞
C
l
−
=
109
......(iii)
On adding Eq. (i) and (ii) and subtracting the Eq. (iii)
Λ
∞
N
H
4
+
+
Λ
∞
O
H
−
=
130
+
217
−
109
Λ
∞
N
H
4
O
H
=
347
−
109
=
238
ohm
−
1
cm
2
equiv
−
1
© examsnet.com
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