Question

A conductivity cell has a cell constant of 0.5 cm^-1. This cell when filled with 0.01 M NaCl solution has a resistance of 384 ohms at 25$^{\circ}$C. Calculate the equivalent conductance of the given solution.

• A. $130.2\, \Omega^{-1} cm^2 (g \,eq)^{-1}$
• B. $137.4 \, \Omega^{-1} cm^2 (g \,eq)^{-1}$
• C. $154.6 \, \Omega^{-1} cm^2 (g \,eq)^{-1}$
• D. $169.2 \, \Omega^{-1} cm^2 (g \,eq)^{-1}$

Answer 1

Answer: (A)

Equivalent conductance, $Λ_{eq}=\frac{k\times1000}{Normality}$
where, κ (Specific conductance) = C $\times\frac{l}{a}$
=$\frac{1}{R}\times\frac{l}{a}=\frac{1}{384}\times0.5$
$= 1.302 × 10^{-3} ohm^{-1} cm^{-1}$
$\therefore \quadΛ_{eq}=\frac{1.302\times10^{-3}\times1000}{0.01}$
= 130.2 ohm$^{-1} cm^{2}$ (g eq)$^{-1}$