Question

A conductivity cell filled with 0.02 M $AgN{{O}_{3}}$ gives at 25 $^{o}C$ a resistance of 947 ohms. If the cell constant is 2.3 $c{{m}^{-1}}$ . What is the molar conductivity of the 0.02 M $AgN{{O}_{3}}$ at 25 $^{o}C$ .
A. 121.43 ${{\Omega }^{-1}}c{{m}^{2}}mol{{e}^{-1}}$
B. 101.4 ${{\Omega }^{-1}}c{{m}^{2}}mol{{e}^{-1}}$
C. 111.4 ${{\Omega }^{-1}}c{{m}^{2}}mol{{e}^{-1}}$
D. None of the above.

Answer 1

Hint: There is a formula to calculate the molar conductivity of a solution. The formula is as follows.
\[{{\Lambda }_{m}}=\dfrac{R\times 1000}{M}\]
${{\Lambda }_{m}}$ = molar conductivity
R = Specific conductance of the solution
M = Molarity of the solution.

Complete step by step answer:
- In the question they asked to find the molar conductivity of the 0.02 M silver nitrate solution by giving the resistance and molarity of the solution.
- But to calculate the molar conductivity of the solution first we should calculate the specific conductance of the solution by using the following formula.
\[R=G\times \dfrac{l}{a}\]
R = specific conductance of the solution
G = Conductance of the solution = $\dfrac{1}{947}$ (Conductance means reciprocal of Resistance)
$\dfrac{l}{a}$ = Resistance of the solution = 2.3
- Substitute all the known values in the above formula to get the specific conductance of the solution.
\[\begin{align}
  & R=G\times \dfrac{l}{a} \\
 & R=\dfrac{1}{947}\times 2.3 \\
 & R=2.42\times {{10}^{-3}}oh{{m}^{-1}}c{{m}^{-1}} \\
\end{align}\]

- From specific conductance of the solution we can calculate the molar conductivity of the solution by using the following formula.
\[{{\Lambda }_{m}}=\dfrac{R\times 1000}{M}\]
${{\Lambda }_{m}}$ = molar conductivity
R = Specific conductance of the solution = $2.42\times {{10}^{-3}}oh{{m}^{-1}}c{{m}^{-1}}$
M = Molarity of the solution = 0.02 M

- Substitute all the known values in the above formula to get the molar conductivity of the solution
\[\begin{align}
  & {{\Lambda }_{m}}=\dfrac{R\times 1000}{M} \\
 & {{\Lambda }_{m}}=\dfrac{2.42\times {{10}^{-3}}\times 1000}{0.02} \\
 & {{\Lambda }_{m}}=121.43{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}} \\
\end{align}\]
- Therefore the molar conductivity of the solution is 121.43 ${{\Omega }^{-1}}c{{m}^{2}}mol{{e}^{-1}}$ .
So, the correct answer is “Option A”.

Note: Don’t be confused with Siemen, $oh{{m}^{-1}}$ and ${{\Omega }^{-1}}$ . All are the same and going to denote the resistance only. The relation between them is as follows.
Siemen = $oh{{m}^{-1}}$ = ${{\Omega }^{-1}}$ .
The above are the units to represent the resistance in different forms.