Question

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

Answer 1

Hint: Since the value of diameter is given in the question, we can calculate the area of the column. And, the electrical resistance of a column is also given and we know the formula for Electrical resistance, R is given by $\dfrac{{\rho \iota }}{A}$, we can calculate resistivity, $\rho $ from the equation and the conductivity is the reciprocal of the resistivity, \[\rho \]. Further, we can calculate the molar conductivity from the formula, $\dfrac{{1000 \times K}}{C}$.

Complete step by step answer:
Given in the question are,
Diameter = 1 cm
 $ \Rightarrow $ Radius, \[R = \dfrac{D}{2} = \dfrac{1}{2} = 0.5cm\]
Length = 50 cm
We know,
 \[Area = \pi {r^2} = 3.14 \times 0.05 \times 0.05 = 0.785c{m^2}\] (Since, \[\pi = 3.14\] )
Electrical resistance, $R = \dfrac{{\rho \iota }}{A}$
Resistivity, \[\rho = \dfrac{{RA}}{\iota }\]
Where,
R = Resistance
A = area of cross section
l = length
Substituting the values in the formula of Resistivity,
 \[\rho = \dfrac{{RA}}{\iota }\]
So, \[\rho = \dfrac{{5.55 \times {{10}^{ - 3}} \times 0.785}}{{50}} = 87.135\,ohm\,cm\]
Conductivity is the reciprocal of the resistivity of the material
Conductivity, $K = \dfrac{1}{\rho } = \dfrac{1}{{87.135}} = 0.01148\,S\,c{m^{ - 1}}$
Molar conductivity $ = \dfrac{{1000 \times K}}{C} = \dfrac{{0.01147 \times 1000}}{{0.05}}$
Molar conductivity \[ = 229.4\,Sc{m^2}\,mo{l^{ - 1}}\]

Note: The standard unit of conductance is S (Siemens). Specific conductivity or conductivity is the measure of the ability of that material to conduct the electricity. It is represented by the symbol ‘K’. The conductance of the material depends upon the nature of the material. The conductivity of the electrolytic solutions depends upon the following reasons: the nature and the concentration of the electrolyte that is added, the size of the ions that are produced, temperature and the solvent nature and viscosity.