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Electrochemistry question

2022 · 29 Jul · Shift 1 · Q16

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JEE Main
Chemistry
Electrochemistry
2022 · 29 Jul · Shift 1 · Q16

Electrochemistry question

2022 · 29 Jul · Shift 1 · Q16

JEE MainChemistryElectrochemistryNumerical+4 / −1

Resistance of a conductivity cell (cell constant

129 \mathrm{~m}^{-1}

) filled with

74.5 \,\mathrm{ppm}

solution of

\mathrm{KCl}

is

100 \,\Omega

(labelled as solution 1). When the same cell is filled with

\mathrm{KCl}

solution of

149 \,\mathrm{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

\underline{\hspace{2cm}}

. (Nearest integer) Given, molar mass of

\mathrm{KCl}

is

74.5 \mathrm{~g} \mathrm{~mol}^{-1}

.

Numerical answer

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