The resistance of a conductivity cell containing 0.001 M KCl solution…

The correct answer is Option (1) → 0.219 cm^-1

The conductance G of a solution is the reciprocal of its resistance R. Conductivity κ is the conductance of a cell with unit length and unit area of cross-section, so the two are linked by the cell constant G* = l/A:

    \( \displaystyle \kappa=\frac{1}{R}\times\frac{l}{A}=\frac{G^*}{R} \)

    \( \displaystyle G^*=\kappa\times R \)

Here R = 1500 ohm and κ = 0.146 × 10⁻³ S cm⁻¹.

    \( \displaystyle G^*=0.146\times10^{-3}\ \mathrm{S\,cm^{-1}}\times1500\ \mathrm{ohm} \)

    \( \displaystyle G^*=0.219\ \mathrm{cm^{-1}} \)

So the cell constant is 0.219 cm⁻¹.

Shortcut: S × ohm = 1, so κ in S cm⁻¹ multiplied by R in ohm gives the cell constant in cm⁻¹. That rules out Options (3) and (4) at once, and 0.146 × 1.5 = 0.219 settles the value.

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Exercises – “… 2.9 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.”