Question

each of the two platinum electrodes having area ${64mm}^{2}$of a conductivity cells are separated by 8 mm. the resistance of the cell containing $7.5 \times {10}^{3}$M kcl solution at 298K is 1250ohm.calculate cell constant, conductivity,molar conductivity

Answer 1

Hey buddy,

◆ Answer-
a) G = 1.25 /cm
b) K = 0.001 S/cm
c) Λ = 1.33×10^-4 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 = 1.33×10^-4 S/Mcm

Hope that is helpful...

Answer 2

Explanation:

hello please like my answer ❣️