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JEE Main Chemistry Class 11 Electrochemistry Part 2 Questions
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© examsnet.com
Question : 61
Total: 79
Resistance of a conductivity cell (cell constant
129
m
−
1
) filled with
74.5
ppm
solution of
KCl
is
100
Ω
(labelled as solution 1). When the same cell is filled with
KCl
solution of
149
ppm
, the resistance is
50
Ω
(labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e.
∧
1
∧
2
=
x
×
10
−
3
. The value of
x
is _______.(Nearest integer)
Given, molar mass of
KCl
is
74.5
g
mol
−
1
.
[29-Jul-2022-Shift-1]
Your Answer:
Validate
Solution:
l
A
=
129
m
−
1
KCl
solution
1
⇒
74.5
ppm
,
R
1
=
100
Ω
KCl
solution
2
⇒
149
ppm
,
R
2
=
50
Ω
Here,
p
p
m
1
p
p
m
2
=
M
1
M
2
=
(
w
1
∕
M
0
V
×
V
w
2
∕
M
0
)
Λ
1
Λ
2
=
k
1
×
1000
M
1
k
2
×
1000
M
2
=
k
1
k
2
×
M
1
M
2
=
50
100
×
2
=
Λ
1
Λ
2
=
1000
×
10
−
3
=1000
© examsnet.com
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