Question

The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is $5.55\times {10}^3$ ohm. Calculate its resistivity, conductivity, and molar conductivity.

• A. $87.135\times {10}^{-2}ohmm,1.148S{m}^{-1},229.6\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• B. $67.135\times {10}^{-2}ohmm,1.489S{m}^{-1},229.6\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• C. $87.135\times {10}^{-2}ohmm,1.296S{m}^{-1},229.6\times {10}^{-4}S{m}^{2}mo{l}^{-1}$
• D. $77.135\times {10}^{-2}ohmm,1.148S{m}^{-1},229.6\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is A $87.135\times {10}^{-2}ohmm,1.148S{m}^{-1},229.6\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

For resistivity,
We have to find the area.
$Area\left(a\right)=\pi r^2=3.14\times {\left(\frac{1cm}{2}\right)}^2$
$=0.785c{m}^2$
$=0.785\times {10}^{-14}{m}^2$
$I=50cm=0.5m$
Now,
$R=\frac{\rho l}{a}or\rho =\frac{Ra}{l}=\frac{5.55\times {10}^{3}ohm\times 0.758c{m}^2}{50cm}$
$=87.135ohmcm$
$Conductivity\left(k\right)=\frac{1}{\rho }=\left(\frac{1}{87.135}\right)Sc{m}^{-1}$
$=0.01148Sc{m}^{-1}$
$Molarconductivity{\wedge }_m=\frac{k\times 1000}{M}c{m}^3{L}^{-1}$
$=\frac{0.01148Sc{m}^{-1}\times 1000c{m}^3{L}^{-1}}{0.05mol{L}^{-1}}$
$=229.6Sc{m}^{2}mo{l}^{-1}$
Alternatively
The values of different quantities in SI units (i.e., in terms of “m” instead of “cm”)
$\rho =\frac{Ra}{1}=\frac{5.55\times {10}^{3}ohm\times 0.785\times {10}^{-4}{m}^2}{0.5m}$
$=87.135\times {10}^{-2}ohmm$
$k=\frac{1}{\rho }=\frac{100}{87.135}=1.148S{m}^{-1}$
and ${\wedge }_m=\frac{k}{c}=\frac{1.148S{m}^{-1}}{50mol{m}^{-3}}=229.6\times {10}^{-4}S{m}^{2}mo{l}^{-1}$