Question

Resistance of a conductivity cell (cell constant $129\, m ^{-1}$ ) filled with $74.5 \, ppm$ solution of $KCl$ is $100 \, \Omega$ (labelled as solution 1). When the same cell is filled with $KCl$ solution of $149 \, ppm$, 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 $74.5\, g \, mol ^{-1}$

Answer 1

$\frac{\ell}{ A }=129\, m ^{-1}$
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}$