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Question
Chemistry
Following limiting molar conductivities are given as λom(H2SO4) = x S cm2 mol-1 λom(K2SO4) = y S cm2 mol-1 λom(CH3COOK) =z S cm2 mol-1 λom(in S cm2 mol-1) for CH3COOH will be-
Q. Following limiting molar conductivities are given as
λ
m
o
​
(
H
2
​
S
O
4
​
)
=
x
S
c
m
2
m
o
l
−
1
λ
m
o
​
(
K
2
​
S
O
4
​
)
=
y
S
c
m
2
m
o
l
−
1
λ
m
o
​
(
C
H
3
​
COO
K
)
=
z
S
c
m
2
m
o
l
−
1
λ
m
o
​
(
in
S
c
m
2
m
o
l
−
1
)
for
C
H
3
​
COO
H
will be-
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A
x
−
y
+
2
z
15%
B
x
+
y
−
z
12%
C
x
−
y
+
z
15%
D
2
(
x
−
y
)
​
+
z
57%
Solution:
C
H
3
​
COO
H
→
C
H
3
​
CO
O
−
+
H
+
...
(
1
)
H
2
​
S
O
4
​
→
2
H
+
+
S
O
4
−
2
​
...
(
2
)
K
2
​
S
O
4
​
→
2
K
+
+
S
O
4
−
2
​
...
(
3
)
C
H
3
​
COO
K
→
C
H
3
​
CO
O
−
+
K
+
...
(
4
)
According to Kohlrausch's law-
λ
C
H
3
​
COO
H
∘
​
=
λ
C
H
3
​
CO
O
−
∘
​
+
λ
H
+
∘
​
e
q
.
(
1
)
=
e
q
.
(
4
)
+
e
q
.
2
(
2
)
​
−
e
q
.
2
(
3
)
​
∴
λ
C
H
3
​
COO
H
∘
​
=
z
+
2
X
​
−
2
Y
​
λ
C
H
3
​
COO
H
∘
​
=
2
(
X
−
Y
)
​
+
z
(
S
×
c
m
2
m
o
l
−
1
)