The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of 0.88 cm-1. The value of equivalent conductance of solution is -

Question

The resistance of 0.01 N solution of an electrolyte was found to be 220 ohm at 298 K using a conductivity cell with a cell constant of 0.88 cm-1. The value of equivalent conductance of solution is - (A) 400 mho cm2 g eq-1 (B) 295 mho cm2 g eq-1 (C) 419 mho cm2 g eq-1 (D) 425 mho cm2 g eq-1

Tardigrade Admin · Accepted Answer

Lambdaeq = k × (1000/N) = (1/R)× (l/a)× (1000/N) = (1/R) × cell constant × (1000/N) = (1/220)×0.88× (1000/0.01) 400 mho cm2 g eq-1

Q. The resistance of $0.01 N$ solution of an electrolyte was found to be $220 o hm$ at $298 K$ using a conductivity cell with a cell constant of $0.88 c m^{- 1}$ . The value of equivalent conductance of solution is -

$400 mh o c m^{2} g e q^{- 1}$

$295 mh o c m^{2} g e q^{- 1}$

$419 mh o c m^{2} g e q^{- 1}$

$425 mh o c m^{2} g e q^{- 1}$

$Λ_{e q} = k \times \frac{1000}{N} = \frac{1}{R} \times \frac{l}{a} \times \frac{1000}{N}$
$= \frac{1}{R} \times$ cell constant $\times \frac{1000}{N}$
$= \frac{1}{220} \times 0.88 \times \frac{1000}{0.01}$
$400 mh o c m^{2} g e q^{- 1}$

Q. The resistance of 0.01N solution of an electrolyte was found to be 220ohm at 298K using a conductivity cell with a cell constant of 0.88cm−1. The value of equivalent conductance of solution is -

400mhocm2geq−1

295mhocm2geq−1

419mhocm2geq−1

425mhocm2geq−1