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. $130.2{\Omega }^{-1}c{m}^2(geq{)}^{-1}$
• B. $137.4{\Omega }^{-1}c{m}^2(geq{)}^{-1}$
• C. $154.6{\Omega }^{-1}c{m}^2(geq{)}^{-1}$
• D. $169.2{\Omega }^{-1}c{m}^2(geq{)}^{-1}$

Answer 1

Answer: (A)

Equivalent conductance, ${\Lambda }_{eq}=\frac{k\times 1000}{Normality}$
where, κ (Specific conductance) = C $\times \frac{l}{a}$
=$\frac{1}{R}\times \frac{l}{a}=\frac{1}{384}\times 0.5$
$=1.302\times 10^{-3}oh{m}^{-1}c{m}^{-1}$
$\therefore {\Lambda }_{eq}=\frac{1.302\times 10^{-3}\times 1000}{0.01}$
= 130.2 ohm${}^{-1}c{m}^2$ (g eq)${}^{-1}$