Question

Resistance of a $0.1NKCN$ solution is $245\Omega$. If the electrodes in the cell are $5cm$ apart and area having $100c{m}^2$ each, the molar conductance of the solution will be:

• A. $204Sc{m}^{2}mo{l}^{-1}$
• B. $2.04Sc{m}^{2}mo{l}^{-1}$
• C. $20.4Sc{m}^{2}mo{l}^{-1}$
• D. $0.204Sc{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is B $2.04Sc{m}^{2}mo{l}^{-1}$

$\text{Normality,}\left(N\right)=0.1N\text{Resistance,}\left(R\right)=245\Omega \text{Area of cross section,}\left(A\right)=100c{m}^2\text{Length between the electrodes,}\left(l\right)=5cm$

For $KCl$,
n-factor is 1
$\therefore$
$\text{Molarity,}=\frac{1}{10}M$

As we know that,
Molar conductance, $\left({\Lambda }_m\right)=\frac{1000\times \kappa }{M}$
whereas,
$\kappa =\text{Specific conductance}$

$\kappa =\frac{l}{A}\times \frac{1}{R}$

$\therefore {\Lambda }_m=\frac{1000\times \left(\frac{l}{AR}\right)}{M}$

$\Rightarrow {\Lambda }_m=\frac{1000\times 5}{100\times 245\times 0.1}$

$\Rightarrow {\Lambda }_m=2.04Sc{m}^{2}mo{l}^{-1}$