Resistance of a conductivity cell (cell constant 129 m ^ - 1 )...

Given $\dfrac{l}{A} = 129{m}^{- 1}$

$KCl$ solution $1$:

$74 . 5ppm, {R}_{1} = 100Ω$

$KCl$ solution $2$:

$149ppm, {R}_{2} = 50Ω$

Here, $\dfrac{{ppm}_{1}}{{ppm}_{2}} = \dfrac{{M}_{1}}{{M}_{2}}$

and $κ = \dfrac{1}{R}G * whereG * = cellconstant$

Since value of cell constant is same in both the cases.

$\dfrac{{k}_{1}}{{k}_{2}} = \dfrac{{R}_{2}}{{R}_{1}}$

$\dfrac{{∧}_{1}}{{∧}_{2}} = \dfrac{{k}_{1} \times \dfrac{1000}{{M}_{1}}}{{k}_{2} \times \dfrac{1000}{{M}_{2}}}$

$= \dfrac{{K}_{1}}{{K}_{2}} \times \dfrac{{M}_{2}}{{M}_{1}} = \dfrac{{R}_{2}}{{R}_{1}} \times \dfrac{{M}_{2}}{{M}_{1}}$

$= \dfrac{50}{100} \times \dfrac{149}{74 . 5} = 1$

$= \dfrac{{∧}_{1}}{{∧}_{2}} = 1, 000 \times {10}^{- 3}$