Question

Please Answer ASAP!!

=> The resistance of a conductivity cell, containing 0.001 M KCl solution at 298 K. Is 1500 Ohm. What is the cell constant if the conductivity of 0.001 M KCl, solution at 298 K is

0.146 × 10^-3 Scm^-1 ​

Answer 1

${ \bold{ \underline{ \underline{Answer}}}} \implies \: { \sf{Cell \:Constant }} = 0.219 \: cm {}^{ - 1}$

${ \bold{ \underline{ \underline{Step \: by \: Step \: Explanation}}}} \implies$

Using the formula for Cell Constant (G*) :

$\implies \: G {}^{*} \: = { \bold{\dfrac{Conductivity}{Conductance}}}$

Also , we have

$\implies \: Conductance \: (g) \: = \dfrac{1}{Resistance} \\ \\ = >{ \bold{ \sf{ g \: = \dfrac{1}{1500} \: \: \Omega {}^{ - 1}}}}$

$= > { \bold{ \sf{Conductivity \: (k)\: = 0.146 \times 10 {}^{ - 3} S \: cm {}^{ - 1} }}} \: (given)$

➡Plug the values of Conductance and Conductivity into the formula ,

$\implies \: G {}^{*} = \dfrac{0.146 \times 10 {}^{ - 3} }{ \dfrac{1}{1500} } \\ \\ \implies \: G {}^{*} \: = 0.219$

Hence , Cell Constant is

${ \boxed{ \boxed{ \bold{G* = 0.219 \: cm {}^{ - 1} }}}}$

Answer 2

Answer:

see the attachment okay