Question

Resistance of a $0.1$M $KCl$ solution in a conductance cell is $300$ ohm and specific conductance of $0.1$M $KCl$ is $1.29\times {10}^{-2}$ $oh{m}^{-1}$ $c{m}^{-1}$. The resistance of $0.1$M $NaCl$ solution in the same cell is $380$ ohm. Calculate the equivalent conductance of the $0.1$M $NaCl$ solution.

Answer 1

K (specific conductance) $=\frac{1}{R}\left(\frac{1}{A}\right)$

$\frac{l}{A}=1.29\times {10}^{-2}\times 300=3.87$

$l/A=$ cell constant ( cell constant same in both)

In Nacl $So{l}^n$ $k=\frac{1}{380}\left(\frac{l}{A}\right)$

$k=\frac{1}{380}\times 3.87=\frac{3.87}{380}=0.0101$ $on{m}^{-1}c{m}^2$