Question

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

Answer 1

Hint: To solve this question you should know about resistance, conductance. The opposite force to the current is called resistance. Conductance is a reciprocal of resistance of the system. It is also a ratio of the current to the potential difference. The unit of conductance is siemens or \[oh{m^{ - 1}}\].

Step by step solution:
According to the ohm’s law of resistance, resistance is directly proportional of the length and inversely proportional to the cross sectional area. which can be written as, \[R{\text{ }}\alpha {\text{ }}\dfrac{l}{A}\]. Where length is \[{\text{l}}\] and cross sectional area is \[{\text{A}}\] . It can also be written with a proportional constant \[{{\rho }}\] which is called resistivity . Therefore, the formula of resistance is \[R{\text{ }} = {\text{ }}\rho \dfrac{l}{A}\] .
Conductance is a reciprocal of resistance of the system, \[G = \dfrac{1}{R}\]where G is conductance. There for the formula of conductance can be written as, \[G = \dfrac{1}{R} = \dfrac{1}{\rho } \times \dfrac{A}{l}\]. Where\[\dfrac{1}{\rho }\] can be written with a new term specific conductance or conductivity. Therefore, the formula of conductance is,
 \[
G = \kappa \times \dfrac{A}{l} \\
or,G \times \dfrac{l}{A} = \kappa \\
  \]
Now molar conductivity is the ratio of conductivity to molar concentration of the electrolyte.
The formula is, \[G = \dfrac{1}{R}\]
Now the given data is \[R = 5.55{\text{ }} \times {10^3}ohm\],\[l = 50cm\] , \[C = 0.05M\]
 \[
A = \pi {r^2} \\
A = 3.14 \times {\left( {\dfrac{1}{2}} \right)^2} \\
A = 0.785{\text{ }}c{m^2} \\
\]
Now use these values to calculate the resistivity \[\rho \] as follows.
\[
R{\text{ }} = {\text{ }}\rho \dfrac{l}{A} \\
or,5.55{\text{ }} \times {10^3} = {\text{ }}\rho \dfrac{{50}}{{0.785}} \\
or,\dfrac{{0.785}}{{50}} \times 5.55{\text{ }} \times {10^3} = \rho \\
or,0.087 \times {10^3} = \rho \\
\]
The value of resistivity is \[\rho = 0.087 \times {10^3}\Omega cm\].
Now the conductivity is equals to,
\[
\kappa = \dfrac{1}{\rho } \\
= \dfrac{1}{{0.087 \times {{10}^3}\Omega cm}} \\
= 0.01147{\text{ }}Sc{m^{ - 1}} \\
\]
Now the molar conductivity is,
\[
{\Lambda _m} = \dfrac{{1000 \times \kappa }}{c} \\
= \dfrac{{1000 \times 0.01147}}{{0.05}} \\
= 229.4{\text{ }}Sc{m^2}mo{l^{ - 1}} \\
  \]

Note: With dilution of the electrolyte the equivalent conductance and molar conductance value increases. This is because on dilution the degree of dissociation of electrolyte increases as well as the number of ions increases. Remember the definitions of resistance, conductance. Remember the formula of conductance and resistance. The ratio of length to cross-sectional area is called cell constant. Remember the formula of molar conductivity \[{\Lambda _m} = \dfrac{{1000 \times \kappa }}{c}\].