Resistance of decimolar solution is 50 ohm. If electrodes of surface area 0.0004 m^2 each are placed at a distance of 0.02 m then conductivity of solution is : A. 1 s cm^-1 B. 0.01 s cm^-1 C. 0.001 s cm^-1 D. 10 s cm^-1
Added by Valerie T.
Step 1
We are given that R = 50 ohm, L = 0.02 m, and A = 0.0004 m^2. We need to find the conductivity, which is the reciprocal of resistivity, i.e., Ļ = 1/Ļ. Show moreā¦
Show all steps
Your feedback will help us improve your experience
Adi S and 95 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Resistance of a decimolar solution between two electrodes $0.02$ meter apart and $0.0004 \mathrm{~m}^{2}$ in $\cdots$ area was found to be 50 ohm. Specific conductance $(\kappa)$ is : (a) $0.1 \mathrm{Sm}^{-1}$ (b) $1 \mathrm{~S} \mathrm{~m}^{-1}$ (c) $10 \mathrm{~S} \mathrm{~m}^{-1}$ (d) $4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^{-1}$
Resistance of $0.1 \mathrm{M} \mathrm{KCl}$ solution in a conductance cell is $300 \mathrm{ohm}$ and conductivity is $0.013 \mathrm{Scm}^{-1}$. The value of cell constant is: (a) $3.9 \mathrm{~cm}^{-1}$ (b) $39 \mathrm{~m}^{-1}$ (c) $3.9 \mathrm{~m}^{-1}$ (d) None of these
A conductance cell was filled with a $0.02 \mathrm{M} \mathrm{KCl}$ solution which has a specific conductance of $2.768 \times 10^{-3} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$. If its resistance is $82.4 \mathrm{ohm}$ at $25^{\circ} \mathrm{C}$, the cell constant is : (a) $0.2182 \mathrm{~cm}^{-1}$ (b) $0.2281 \mathrm{~cm}^{-1}$ (c) $0.2821 \mathrm{~cm}^{-1}$ (d) $0.2381 \mathrm{~cm}^{-1}$
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD