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JEE Main Chemistry Class 11 Electrochemistry Part 2 Questions
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© examsnet.com
Question : 66
Total: 79
The resistance of a conductivity cell containing
0.01
M
KCl
solution at
298
K
is
1750
Ω
. If the conductivity of
0.01
M
KCl
solution at
298
K
is
0.152
×
10
−
3
S
cm
−
1
, then the cell constant of the conductivity cell is _____
×
10
−
3
cm
−
1
[24-Jun-2022-Shift-2]
Your Answer:
Validate
Solution:
Molarity of
KCl
solution
=
0.1
M
Resistance
=
1750
ohm
Conductivity
=
0.152
×
10
−
3
S
cm
−
1
Conductivity
=
Cell constant
Resistance
∴
Cell constant
=
0.152
×
10
−
3
×
1750
=
266
×
10
−
3
cm
−
1
© examsnet.com
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