Electrochemistry - Result Question 13
####14. If $0.01 M$ solution of an electrolyte has a resistance of $40 ohms$ in a cell having a cell constant of $0.4 cm^{-1}$, then its molar conductance in ohm ${ }^{-1} cm^{2} mol^{-1}$ is
(a) $10^{2}$
(b) $10^{4}$
(c) 10
(d) $10^{3}$
[1997]
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Solution:
- (d) Given molarity $=0.01 M$
Resistance $=40$ ohm;
Cell constant $\frac{1}{A}=0.4 cm^{-1}$
Specific conductivity ( $\kappa$ )
$=\frac{\text{ cell constant }}{\text{ resistance }}=\frac{0.4}{40}=0.01 ohm^{-1} cm^{-1}$
Molar conductance $(\wedge_m)=\frac{1000 \kappa}{\text{ Molarity }}$
$=\frac{1000 \times .01}{.01}=10^{3} ohm^{-1} cm^{2} mol^{-1}$
