Question

The electrical resistance of a column of $0.05 \,mol \,L^{-1} NaOH$ solution of diameter $1 \,cm$ and length $50 \,cm$ is $5.55 \times 10^3$ ohm. The conductivity is

• A. $0.01148\,S\, cm^{-1}$
• B. $0.00052 \,S\, m^{-1}$
• C. $154.5 \,S \,cm^{-1}$
• D. $5.075\, S \,m^{-1}$

Answer 1

Answer: (A)

$1 = 50 \,cm = 0.5 \,m$
$A = \pi r^2 = 3.14 \times (0.5)^2 cm^2 $
$ = 0.785 \,cm^2 $
$= 0.785 \times 10^{-4} m^2$
$R = \rho \frac{l}{A}$ or $\rho = \frac{RA}{l} $
$= \frac{5.55 \times 10^3 \,\Omega \times 0.785 \,cm^2}{50\,cm} $
$ = 87.135\,\Omega\,cm$
Conductivity $= K \frac{1}{\rho} = \left(\frac{1}{87.135}\right)S \,cm^{-1}$
$ = 0.01148\,S\,cm^{-1}$