The resistance of a conductivity cell containing $0 . 01MKCl$ solution at $298K$ is $1750Ω$ . If the conductivity of $0 . 01MKCl$ solution at $298K$
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 . 01MKCl

solution at

298K

is

1750Ω

. If the conductivity of

0 . 01MKCl

solution at

298K

is

0 . 152 \times {10}^{- 3}S{cm}^{- 1}

, then the cell constant of the conductivity cell is____

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

Quantrex Academy · Accepted Answer

Correct answer: 0

For KCl solution, R = 1750 ohms Conductivity, K = 0 . 152 × 10 ^ - 3 S cm ^ - 1 K=((l)/(a))(1)/(R) Cell constant, (l)/(a) = KXR=0 . 152 × 10 ^ - 3 X1750 = 266X 10 ^ - 3 cm ^ - 1

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