Question

The resistance of $0.01NNaCl$ solution at ${25}^{\circ }C$ is $200\Omega$. Cell constant of conductivity cell is $1c{m}^{-1}$. The equivalent conductance is :

• A. $5\times {10}^2{\Omega }^{-1}{\text{cm}}^2(\text{g.eq}{)}^{-1}$
• B. $6\times {10}^3{\Omega }^{-1}{\text{cm}}^2(\text{g.eq}{)}^{-1}$
• C. $7\times {10}^4{\Omega }^{-1}{\text{cm}}^2(\text{g.eq}{)}^{-1}$
• D. $8\times {10}^5{\Omega }^{-1}{\text{cm}}^2(\text{g.eq}{)}^{-1}$

Answer 1

The correct option is A $5\times {10}^2{\Omega }^{-1}{\text{cm}}^2(\text{g.eq}{)}^{-1}$

${\lambda }_{eq}=\frac{k\times 1000}{N}=\frac{Cx\times 1000}{N}$
$=\frac{1}{R}\times \frac{l}{a}\times \frac{1000}{N}$
${\lambda }_{eq}=\frac{1c{m}^{-1}}{200\Omega }\times \frac{1000}{0.01N}=5\times {10}^2{\Omega }^{-1}c{m}^2(g.eq{)}^{-1}$