Question

The resistance of 0.1 N solution of a salt is found to be $2.5\times {10}^3$. The equivalent conductance of the solution is: (cell constant=1.15 $c{m}^{-1}$)

• A. 3.6
• B. 4.6
• C. 5.6
• D. 6.6

Answer 1

Answer: (B)

4.6

The relationship between the specific conductance, resistance and the cell constant is $\kappa =\frac{1}{R}\times \frac{l}{a}$.
Substitute $R=2.5\times {10}^{3}ohm$ and $\frac{l}{a}=1.15{cm}^{-1}$.
Hence $\kappa =\frac{1}{2.5\times {10}^{3}ohm}\times 1.15{cm}^{-1}=\frac{1.15}{2.5\times {10}^3}{ohm}^{-1}{cm}^{-1}$.
The relationship between the equivalent conductance and specifc conductance is ${\Lambda }_{eq}=\frac{\kappa \times 1000}{M}$.
Substitute $M=0.1N$ and $\kappa =\frac{1.15}{2.5\times {10}^3}{ohm}^{-1}{cm}^{-1}$.
Hence ${\Lambda }_{eq}=\frac{\frac{1.15}{2.5\times {10}^3}\times 1000}{0.1}=4.6{ohm}^{-1}{cm}^2{equiv}^{-1}$.