Resistance of a conductivity cell (cell constant $129 \mathrm{~m}^{-1}$) 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 $74.5 \mathrm{~g} \mathrm{~mol}^{-1}$

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Redox Reactions and Electrochemistry

Resistance of a conductivity cell (cell constant

129 m^{- 1}

) filled with 74.5 ppm solution of KCl is

100Ω

(labelled as solution 1). When the same cell is filled with KCl solution of 149 ppm, the resistance is

50Ω

(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 1 0^{- 3}

. The value of

x

is ______.
(Nearest integer)
Given, molar mass of KCl is

74.5 g mol^{- 1}

Answer: 1000

The molar conductivity ratio is given by:

\frac{\land _{1}}{\land _{2}} = \frac{κ _{1}}{κ _{2}} \times \frac{M _{2}}{M _{1}}

Since conductivity

κ = \frac{cell constant}{R}

\frac{κ _{1}}{κ _{2}} = \frac{R _{2}}{R _{1}} = \frac{50}{100} = \frac{1}{2}

For ppm solutions:

\frac{ppm _{1}}{ppm _{2}} = \frac{M _{1}}{M _{2}}

, so

\frac{M _{2}}{M _{1}} = \frac{149}{74.5} = 2

Therefore:

\frac{\land _{1}}{\land _{2}} = \frac{1}{2} \times 2 = 1 = 1000 \times 1 0^{- 3}

Hence

x = 1000

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer