The resistance of 0.01 m KCl solution at 298 K is 1500 Ω. If the conductivity of 0.01 m KCl solution at 298 K is 0.146 × 10-3 S cm -1 . The cell constant of the conductivity cell in cm -1 is

Question

The resistance of 0.01 m KCl solution at 298 K is 1500 Ω. If the conductivity of 0.01 m KCl solution at 298 K is 0.146 × 10-3 S cm -1 . The cell constant of the conductivity cell in cm -1 is (A) 0.219 (B) 0.291 (C) 0.301 (D) 0.194

Tardigrade Admin · Accepted Answer

kappa=(1/ R )((l/ a )) ∴ (l/ a )= kappa ⋅ R =0.146 × 10-3 × 1500 =219 × 103=0.219 cm -1

Q. The resistance of $0.01 m K Cl$ solution at $298 K$ is $1500 Ω$ . If the conductivity of $0.01 m K Cl$ solution at $298 K$ is $0.146 \times 1 0^{- 3} S c m^{- 1} .$ The cell constant of the conductivity cell in $c m^{- 1}$ is

0.219

0.291

0.301

0.194

$κ = \frac{1}{R} (\frac{l}{a})$
$∴ \frac{l}{a} = κ \cdot R$
$= 0.146 \times 1 0^{- 3} \times 1500$
$= 219 \times 1 0^{3} = 0.219 c m^{- 1}$

Q. The resistance of 0.01mKCl solution at 298K is 1500Ω. If the conductivity of 0.01mKCl solution at 298K is 0.146×10−3Scm−1. The cell constant of the conductivity cell in cm−1 is