A KCl solution of conductivity 0.14 Sm⁻¹ shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω. The conductivity of the HCl solution is ______ ×10⁻²S m⁻¹.
A KCl solution of conductivity 0.14 Sm⁻¹ shows a resistance of 4.19 Ω in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03 Ω. The conductivity of the HCl solution is ______ ×10⁻²S m⁻¹.
Asked by Shiksha User
-
1 Answer
-
The cell constant (G) is given by G = κ × R, where κ is conductivity and R is resistance.
Since the cell constant is constant: (κ × R)KCl = (κ × R)HCl
0.14 Sm? ¹ × 4.19 Ω = κ_HCl × 1.03 Ω
κ_HCl = (0.14 × 4.19) / 1.03 = 0.569 Sm? ¹
This is equivalent to 56.9 × 10? ² Sm? ¹.
The answer, rounded off, is 57.
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers