wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A KCl solution of conductivity 0.14 Sm1 shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops of 1.03 Ω. The conductivity of the HCI solution is ____ ×102 S m1. (Round off to the Nearest Integer).

A
57.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
57
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
57.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

For KCl

k1=0.14 Sm1

R1=4.19Ω

For HCl

k2=x Sm1 (Let)

R2=1.03 Ω

R=ρ(la)kR=(la)Cell constant

k1R1=k2R2

0.14×4.19=1.03×k2

k2=0.5695 Sm1

k2=57×102Sm1

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Faraday's Laws
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon