Question

Resistance of a conductivity cell (cell constant $129m^{-1}$ ) filled with $74.5ppm$ solution of $KCl$ is $100\Omega$ (labelled as solution 1). When the same cell is filled with $KCl$ solution of $149ppm$, the resistance is $50\Omega$ (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e. $\frac{{\wedge }_1}{{\wedge }_2}=x\times 10^{-3}$. The value of $x$ is___ (Nearest integer)
Given, molar mass of $KCl$ is $74.5gmo{l}^{-1}$

Answer 1

Answer: 1000

$\frac{\ell }{A}=129m^{-1}$
KCl solution 1 :
$74.5ppm,R_1=100\Omega$
$KCl$ solution 2 :
$149ppm,R_2=50\Omega$
$149ppm,R_2=50\Omega$
Here, $\frac{pp{m}_1}{pp{m}_2}=\frac{M_1}{M_2}\left(=\frac{W_1/M_0}{V}\times \frac{V}{W_2/M_0}\right)$
$\frac{{\wedge }_1}{{\wedge }_2}=\frac{{\kappa }_1\times \frac{1000}{M_1}}{{\kappa }_2\times \frac{1000}{M_2}}$
$=\frac{{\kappa }_1}{{\kappa }_2}\times \frac{M_2}{M_1}$
$=\frac{50}{100}\times 2$
$=\frac{{\wedge }_1}{{\wedge }_2}=1,000\times 10^{-3}$