Ncert Exercises & Solutions

3.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1.

Hint: cell constant = $\frac{k}{Conductance}$

Step 1:

The formula is as follows:

cell constant = $\frac{k}{Conductance}$

Given values are as follows:

Conductivity, $k = 0.146 \times 10^{- 3} S {cm}^{- 1}$

Step 2:

Resistance, R=1500 $Ω$

$∴$ Cell constant=k x R

= $0.146 \times 10^{- 3} \times 1500$

$= 0.219 {cm}^{- 1}$

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