Resistance of a conductivity cell (Cell constant 129 m^–1) filled with 74.5 ppm solution of KCl is $100 Ω$ (labelled as solution 1). When the same cell is filled with KCl solution of 19 ppm, the resistance is $50 Ω$ (labelled as solution 2). The ratio of molar conductivity of solution 1 and solution 2 is i.e., $\frac{\land_{1}}{\land_{2}} = x \times 10^{- 3} .$ The value of x is (nearest integer)
Given, molar mass of KCl is $74.5 g m o l^{- 1}$

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Detailed Solution

$\frac{ℓ}{A} = 129 m^{- 1}$

KCl solution 1

74.5 ppm, $R_{1} = 100 Ω$ , M₁= 1x 10 ^{- 3}, cell constant=129

KCl solution 2

149 ppm, $R_{2} = 50 Ω$ , M₂= 2 x 10 ^{- 3}, cell constant=129

Here, $\frac{p p m_{1}}{p p m_{2}} = \frac{M_{1}}{M_{2}} (\frac{W_{1} / W_{2}}{V} \times \frac{V}{W_{2} / M_{0}})$

$\frac{\land_{1}}{\land_{2}} = \frac{κ_{1} \times \frac{1000}{M_{1}}}{κ_{2} \times \frac{1000}{M_{2}}}$

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

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

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

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