Question

The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 S cm−1.

Answer 1

Answer:

0.219 cm−1

Explanation:

Conductivity of the cell = K = 0.146 × 10−3 S cm−1  (Given)

Resistance of the cell , R = 1500 Ω  (Given)

Conductivity of solution = 0.001M KCl (Given)

According to the formula of cell constant -  

Cell constant = κ × R

Substituting the values we will get -  

= 0.146 × 10−3 × 1500

= 0.219

Therefore, the cell constant if conductivity of 0.001M KCl is 0.219 cm−1

Answer 2

Answer:

see the attachment okay