Engineering Chemistry S Datta

220 ENGINEERING CHEMISTRY
Example 3. Specific conductivity of a conductivity cell containing 0.2 (N) KCl is 2.767 ×
10 –3 ohm –1 cm –1 , the observed resistance of the cell is 4364 ohm. Calculate the cell constant. If
the cell is filled with another solution the observed resistance is 3050 ohm. Calculate specific
resistance of the solution.
Sol.
Cell constant = l a R = ρ l a
or l a = R ρ = R × 1 ρ = R × K
∴ l a = R × K = 4364 × 2.767 × 10–3 cm –1 = 12.075 cm –1
ρ = R
la /
= 3050
12.
075
ohm. cm.
= 252.58 ohm. cm.
Example 4. A conductivity cell containing 0.02 N KCl solution offers a resistance of
550 ohm at a definite temperature. The specific conductivity of 0.02 N solution at that temperature
is 0.002768 ohm –1 cm –1 . If the cell is now filled with 0.2 N ZnSO 4
, the observed resistance of the
cell is 72.18 ohm. Calculate the equivalent and molar conductivities of the ZnSO 4
solution.
Sol. Cell constant = R × K
= 550 × 0.002768 = 1.5224 cm –1
Again, K = cell constant × observed conductivity
= 1.5224 ×
1
72.
18
λ v
(equivalent conductivity) = K v
× 1000
C
λ m
(molar conductivity) = 1000K
M
= 0.02109 ohm –1 cm –1
= 0.02109 × 1000
0.2
= 105.45 ohm –1 cm 2
1000 × 0.02109
=
0.1
= 210.9 ohm –1 cm 2 .
Example 5. The resistance of a decinormal solution of a salt occupying a volume between
two Pt electrodes 1.80 cm apart and 5.4 cm 2 in area was found to be 32 ohms. Calculate the
equivalent conductance (λ v
) of the solution.
Sol.
R = ρ l a
or
1
ρ = 1 R
l
a = 1
32
18
× . 54 .
K = 0.010416 ohm –1 cm –2
λ v
= 1000K
C
=
1000 × 0.010416
0.1
= 104.166 ohm –1 cm –2 .