Question

The resistance of $0.01\text{N}NaCl$ solution at $25{}^{\circ }\text{C}$ is $200\Omega$. Cell constant of conductivity cell is $1{\text{cm}}^{-1}$. 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 $5\times {10}^2{\Omega }^{-1}{\text{cm}}^2(\text{g.eq}{)}^{-1}$

According to the question
Concentration = $0.01\text{N}$
Resistance $\left(R\right)=200\Omega$
Cell constant $\left(\frac{l}{a}\right)=1{\text{cm}}^{-1}$
Equivalent conductance is given by
${\lambda }_{\text{eq}}=\frac{k\times 1000}{\text{N}}$
$=\frac{1}{\text{R}}\times \frac{l}{a}\times \frac{1000}{\text{N}}(\because k=C\times \frac{l}{a}=\frac{1}{\text{R}}\times \frac{l}{a})$
$\Rightarrow {\lambda }_{\text{eq}}=\frac{1{\text{cm}}^{-1}}{200\Omega }\times \frac{1000}{0.01\text{N}}\Rightarrow {\lambda }_{eq}=5\times {10}^2{\Omega }^{-1}{\text{cm}}^2(\text{g.eq}{)}^{-1}$