Question

Each of the two platinum electrodes having area 64 mm^2 of a conductivity cell are separated by 8 mm.The resistance of the cell containing 7.5×10^-3 KCl solution at 298 K 1250 ohm.Calculate,cell constant,conductivity and molar conductivity.

Answer 1

Hey buddy,

◆ Answer-
a) G = 1.25 /cm
b) K = 0.001 S/cm
c) Λ = 133.33 S/Mcm

◆ Explanation-
# Given-
A = 64 mm^2 = 0.64 cm^2
l = 8 mm = 0.8 cm
R = 1250 Ω
T = 298 K
M = 7.5×10^-3 M

# Formula-
a) Cell constant-
G = l/A
G = 0.8/0.64
G = 1.25 /cm

b) Conductivity-
K = G/R
K = 1.25/1250
K = 0.001 S/cm

c) Molar conductivity-
Λ = K/M × 1000
Λ = (0.001×1000) / (7.5×10^-3)
Λ = 133.33 S/Mcm

Hope that is helpful...

Answer 2

Explanation:

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