the resistance of a conductivity cell with 0.1 M KCL solution is 200 ohm when the same cell is fitted with 0.02 M NaCl solution the resistance is 1100 Ohm. Given that the conductivity of 0.1 m KCL solution is 1.29 Ohm-¹m-¹. calculate the cell constant and molar conductivity of 0.02 M NaCl solution
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Answer:
For 0.01 M KCl:
Cell constant = conductivity × resistance
=1.29×100=129m
−1
For 0.02 M KCl:
Cell constant = conductivity × resistance
129=conductivity ×520
∴conductivity=0.248 Sm
−1
=2.48×10
−3
SCm
−1
Also Molar conductivity =conductivity×
M
(
mole/1)
1000
2.48×10
−3
×
M
(
0.02)
1000
=124Scm
2
mol
−1
or molar conducitivity =Conductivity(Sm
−1
)×
M
(mole/m
3
)
1
0.02×10
3
0.248×1
=12.4×10
−3
=124×10
−4
Sm
2
mole
−1
[M=0.02
litre
mole
=0.02
dm
3
mole
=0.02
10
−3
m
3
mole
=0.02×10
3
m
3
mole
]
λ=
N
K×1000
=
0.1
1.11×
10
−2
×1000
111Scm
2
eq
−1
=111×10
4
Scm
2
eq
−1
λ=
N
(eq/m
3
)
K
(Sm
−1
)
=
0.1×10
3
1.11
111×10
4
Scm
2
eq
−1
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