The resistance of a conductivity cell containing $0.01 MKCl$ solution at $298 K$ is $1750 Ω$ . If the conductivity of $0.01 MKCl$ solution at $298 K$
is $0.152 \times 10^{- 3} S {cm}^{- 1}$ , then the cell constant of the conductivity cell is____ $\times 10^{- 3} {cm}^{- 1}$

Question

The resistance of a conductivity cell containing

0.01 MKCl

solution at

298 K

is

1750 Ω

. If the conductivity of

0.01 MKCl

solution at

298 K

is

0.152 \times 10^{- 3} S {cm}^{- 1}

, then the cell constant of the conductivity cell is____

\times 10^{- 3} {cm}^{- 1}

Aicharya · Accepted Answer

For KCl solution, R = 1750 ohms

Conductivity, K = $0.152 \times 10^{- 3} S {cm}^{- 1}$

K= $(\frac{l}{a}) \frac{1}{R}$

Cell constant, $\frac{l}{a} = K XR$ = $0.152 \times 10^{- 3} X 1750 = 266 X 10^{- 3} {cm}^{- 1}$

The resistance of a conductivity cell containing 0 . 01   MKCl solution at 298   K is 1750  …