QJEEM21Ch495e: JEE Main 2021 Chemistry - Class 12 JEEM21Ch495e Solution: The resistance of a conductivity cell with cell constant 1.14…

Question

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The resistance of a conductivity cell with cell constant 1.14 cm^{$-$ 1}, containing 0.001 M KCl at 298 K is 1500

Ω

. The molar conductivity of 0.001 M KCl solution at 298 K in S cm² mol^{$-$ 1} is ____________. (Integer answer)

Accepted Answer

K = \frac{1}{R} \times \frac{l}{A} = ((\frac{1}{1500}) \times 1.14)

S cm^{$-$ 1}

\Rightarrow \land_{m} = 1000 \times \frac{(\frac{1.14}{1500})}{0.001}

S cm² mol^{$-$ 1}

= 760 S cm² mol^{$-$ 1}

\Rightarrow

760

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