Question

The resistance of $0.05MNaOH$ solution is $36.1\Omega$ and cell constant is $0.357c{m}^{-1}$. Calculate its molar conductivity.

• A. ${\Lambda }_m=200{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
• B. ${\Lambda }_m=159{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
• C. ${\Lambda }_m=290{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
• D. ${\Lambda }_m=598{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is A ${\Lambda }_m=200{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$

$\text{Cell constant,}{G}^{\ast }=0.357c{m}^{-1}$
$\text{Molar concentration,}\left(C\right)=0.05M$
$\text{Resistance,}\left(R\right)=31.6\Omega$
We know,

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

$\frac{l}{A}$ is the cell constant $\left(G^{\ast }\right)$ of the cell.
$\therefore$

$\kappa =\frac{0.357}{36.1}\approx 0.01$
Thus,
$\text{Conductivity,}\kappa =0.01Sc{m}^{-1}$

Also, we know
$1L=1000c{m}^3$
$1L^{-1}={10}^{-3}c{m}^{-3}$
$\text{Molar concentration,}\left(C\right)=0.05mol{L}^{-1}$
$\text{Molar concentration,}\left(C\right)=0.05\times {10}^{-3}molc{m}^{-3}$
We know,
$\text{Molar conductance,}{\Lambda }_m=\frac{\kappa }{C}$
${\Lambda }_m=\frac{1000\times 0.01}{0.05}$
${\Lambda }_m=200Sc{m}^{2}mo{l}^{-1}$
Thus,
$\text{Molar Conductivity,}{\Lambda }_m=200{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$