Question

The resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 ohm. The conductivity of this solution is 1.29 \[S\;{m^{ - 1}}\]. Resistance of the same cell when filled with 0.2 M of the same solution is 520 ohm. The molar conductivity of 0.02M solution of the electrolyte will be:
A. \[124 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}\]
B. \[1240 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}\]
C. \[1.24 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}\]
D. \[12.4 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}\]

Answer 1

Hint:The conductance is defined as the inverse of resistance. The conductivity of the solution is equal to length divided by area multiplied by conductance. Volume is calculated in \[c{m^3}\]which is related to the molar conductance. The molar conductance is calculated by multiplying conductivity multiplied by volume.

Complete step by step answer:
Given
Resistance for electrolyte containing 0.1M solution is 100 ohm.
Concentration of electrolyte is 0.1M.
The conductivity of the solution is 1.29 \[S\;{m^{ - 1}}\].
The relationship between the conductivity and resistance is shown below.
Resistance for a cell containing 0.2 M solution is 520 ohm.
\[K = \dfrac{1}{R}\left( {\dfrac{l}{a}} \right)\]
Where,
K is the conductivity
R is the resistance
l is the length
a is the area.
Substitute the values in the above equation.
\[ \Rightarrow 1.29 = \dfrac{1}{{100}}\left( {\dfrac{l}{a}} \right)\]
\[ \Rightarrow \left( {\dfrac{l}{a}} \right) = 129{m^{ - 1}}\]
For 0.2 M solution
\[ \Rightarrow k = \dfrac{1}{{520}}\left( {129} \right){\Omega ^{ - 1}}{m^{ - 1}}\]
The relation between the molar conductance and conductivity is shown below.
The formula to calculate the molar conductance is shown below.
\[\mu = k \times V\]
Where,
\[\mu \] is molar conductance
K is the conductivity.
V is the volume.
Substitute the values in the above equation.
\[ \Rightarrow \mu = \dfrac{1}{{520}} \times 129 \times \dfrac{{1000}}{{0.02}} \times {10^{ - 6}}{m^3}\]
\[ \Rightarrow \mu = \dfrac{{129}}{{520}} \times \dfrac{{1000}}{{0.02}} \times {10^{ - 6}}{m^3}\]
\[ \Rightarrow \mu = 124 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}\]
Thus, the molar conductivity of the solution is \[124 \times {10^{ - 4}}S{m^2}mo{l^{ - 1}}\].
Therefore, the correct option is A.

Note:
The resistance of the conductor varies directly to its length (l) and inversely to the cross sectional area (A). Make sure to convert the value in \[c{m^3}\]. l = 1 cm and A = 1 \[c{m^2}\].