Resistance of a conductivity cell (cell constant 129   m - 1 ) filled with 74 . 5 ppm solution of KCl…

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

$\mathrm{KCl}$ solution $1$:

$74.5\mathrm{ppm}, {\mathrm{R}}_1=100 \mathrm{\Omega }$

$\mathrm{KCl}$ solution $2$:

$149\mathrm{ppm}, {\mathrm{R}}_2=50 \mathrm{\Omega }$

Here, $\frac{{\mathrm{ppm}}_1}{{\mathrm{ppm}}_2}=\frac{{\mathrm{M}}_1}{{\mathrm{M}}_2}$

and $\mathrm{\kappa } = \frac{1}{\mathrm{R}}\mathrm{G}* \mathrm{where} \mathrm{G}* = \mathrm{cell} \mathrm{constant}$

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

$\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2} = \frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}$

$\frac{{\wedge }_1}{{\wedge }_2}=\frac{{\mathrm{k}}_1\times \frac{1000}{{\mathrm{M}}_1}}{{\mathrm{k}}_2\times \frac{1000}{{\mathrm{M}}_2}}$

$=\frac{{\mathrm{K}}_1}{ {\mathrm{K}}_2}\times \frac{{\mathrm{M}}_2}{{\mathrm{M}}_1} = \frac{{\mathrm{R}}_2}{ {\mathrm{R}}_1}\times \frac{{\mathrm{M}}_2}{{\mathrm{M}}_1}$

$=\frac{50}{100}\times \frac{149}{74.5} = 1$

$=\frac{{\wedge }_1}{{\wedge }_2}=1,000\times {10}^{-3}$