$KCl$ solution $1$ :

$74.5\; ppm , R _{1}=100\; \Omega$

$KCl$ solution $2$ :

$149\; ppm , R _{2}=50 \;\Omega$

$149\;ppm , R _{2}=50 \;\Omega$

Here, $\frac{ ppm _{1}}{ ppm _{2}}=\frac{ M _{1}}{ M _{2}}\left(=\frac{ W _{1} / M _{0}}{ V } \times \frac{ V }{ W _{2} / M _{0}}\right)$

$\frac{\wedge_{1}}{\wedge_{2}}=\frac{\kappa_{1} \times \frac{1000}{M_{1}}}{\kappa_{2} \times \frac{1000}{M_{2}}}$

$=\frac{\kappa_{1}}{\kappa_{2}} \times \frac{M_{2}}{M_{1}}$

$=\frac{50}{100} \times 2$

$=\frac{\wedge_{1}}{\wedge_{2}}=1,000 \times 10^{-3}$