Question

$0.1M$ solution is present in a conductivity cell with electrode of $100c{m}^2$ area placed $1cm$ apart and resistance observed was $5\times {10}^{3}ohm$. What is the molar conductivity of this solution?

• A. $5\times {10}^{2}Sc{m}^{2}mo{l}^{-1}$
• B. ${10}^{4}Sc{m}^{2}mo{l}^{-1}$
• C. $200Sc{m}^{2}mo{l}^{-1}$
• D. $0.02Sc{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is D $0.02Sc{m}^{2}mo{l}^{-1}$

Given:
$\text{Resistance,}\left(R\right)=5\times {10}^3\Omega$
$\text{Cell constant,}{G}^{\ast }=\frac{l}{A}$

$\text{Cell constant,}{G}^{\ast }=\frac{1}{100}$

$\text{Cell constant,}{G}^{\ast }={10}^{-2}c{m}^{-1}$

$\text{Conductivity,}\kappa =\frac{1}{R}\times \text{Cell constant}$
$\text{Conductivity,}\kappa =\frac{1}{5\times {10}^3}\times {10}^{-2}$

$\kappa =2\times {10}^{-6}Sc{m}^{-1}$

$\text{Molar concentration,}\left(C\right)=0.1M$

Molar conductivity is the conductance of an ionic solution containing $1mol$ of electrolyte between the two electrodes at a given concentration.
Molar conductivity denoted by the symbol, ${\Lambda }_m.$

If specific conductance ($\kappa$) is the conductance of $1c{m}^3$ of the solution. So, for $Vc{m}^3$ solution,
$\text{Molar conductivity}=\kappa V$
$\kappa$ : Specific conductance / Conductivity
$V$ : Volume of the solution having one mole of an electrolyte

Let,
Molarity of solution is $C$
If $Cmol$ electrolyte dissolves in $1L$ solution, then $1mol$ electrolyte will dissolve in $\frac{1}{C}L$
$\text{Molar conductance,}{\Lambda }_m=\frac{\kappa }{C}$

Also, we know
$1L=1000c{m}^3$
$1L^{-1}={10}^{-3}c{m}^{-3}$
$\text{Molar concentration,}\left(C\right)=0.1mol{L}^{-1}$
$\text{Molar concentration,}\left(C\right)=0.1\times {10}^{-3}molc{m}^{-3}$

${\Lambda }_m=\frac{2\times {10}^{-6}}{0.1\times {10}^{-3}}$

${\Lambda }_m=\frac{1000\times 2\times {10}^{-6}}{0.1}$

${\Lambda }_m=0.02Sc{m}^{2}mo{l}^{-1}$