Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Resistance of 0.1 M KCl solution in a conductance cell is 300 ohm and conductivity is 0.013 Scm-1. The value of cell constant is :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

3.9 cm-1

b

39 cm-1

c

3.9 m-1

d

None of these

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

To determine the cell constant (denoted as K) of a conductance cell, we use the relationship between conductivity (κ), resistance (R), and the cell constant:

Formula: 

K = κ × R

Given Data:

  • Conductivity, κ = 0.013 S/cm
  • Resistance, R = 300 Ω

Conductivity is given as 0.013 S/cm. Convert it to S/m for consistency with SI units: 

0.013 S/cm = 1.3 S/m

Using the formula: 

K = κ × R 

Substitute the values: 

K = 1.3 S/m × 300 Ω = 390 m⁻¹

Since 1 m⁻¹ = 0.01 cm⁻¹, convert the result: 

K = 390 m⁻¹ × 0.01 = 3.9 cm⁻¹

Final Answer:

The cell constant is:  K = 3.9 cm⁻¹

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring