Q. The equivalent conductivity of a solution containing 2.54 g of ${CuSO}_{4}\ per\ L\ is\ 91.0\ \Omega^{- 1}\ {cm}^{2}\ {eq}^{- 1}.$ Its conductivity would be
A) $2.9\ \times 10^{- 3}\ \Omega^{- 1}\ {cm}^{2}\$
B) $1.8\ \times 10^{- 2}\ \Omega^{- 1}\ {cm}^{2}$
C) $2.4\ \times 10^{- 4}\ \Omega^{- 1}\ {cm}^{2}$
D) $3.6\ \times 10^{- 3}\ \Omega^{- 1}\ {cm}^{2}$
Explanation
(a)
$\kappa = \Lambda_{eq} \bullet C$
$= \left( 91\ \Omega^{- 1}\ {cm}^{2}\ {eq}^{- 1} \right)\ \left( \frac{2.54}{159/2 \times 1000}\ eq \bullet {cm}^{- 3} \right)$
$\ = 2.9{\times 10}^{- 3}\ \Omega^{- 1}\ {cm}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.