What would be the equivalent conductivity of a cell in which 0.5 N salt solution offers a resistance of 40 ohm whose electrodes are 2 cm apart and 5 cm2 in area?

Question

What would be the equivalent conductivity of a cell in which 0.5 N salt solution offers a resistance of 40 ohm whose electrodes are 2 cm apart and 5 cm2 in area? (A) 10 ohm-1 cm2 eq-1 (B) 20 ohm-1 cm2 eq-1 (C) 30 ohm-1 cm2 eq-1 (D) 25 ohm-1 cm2 eq-1

Tardigrade Admin · Accepted Answer

kappa=(1/R)×(l/A)=(1/40)×(2/5) Lambdaeq= kappa×(1000/N) =(1/40)×(2/5)×(1000/0.5) =20 ohm-1 cm2 eq-1

Q. What would be the equivalent conductivity of a cell in which $0.5 N$ salt solution offers a resistance of $40 o hm$ whose electrodes are $2 c m$ apart and $5 c m^{2}$ in area?

$10 o h m^{- 1} c m^{2} e q^{- 1}$

$20 o h m^{- 1} c m^{2} e q^{- 1}$

$30 o h m^{- 1} c m^{2} e q^{- 1}$

$25 o h m^{- 1} c m^{2} e q^{- 1}$

$κ = \frac{1}{R} \times \frac{l}{A} = \frac{1}{40} \times \frac{2}{5}$
$Λ_{e q} = κ \times \frac{1000}{N}$
$= \frac{1}{40} \times \frac{2}{5} \times \frac{1000}{0.5}$
$= 20 o h m^{- 1} c m^{2} e q^{- 1}$

Q. What would be the equivalent conductivity of a cell in which 0.5N salt solution offers a resistance of 40ohm whose electrodes are 2cm apart and 5cm2 in area?

10ohm−1cm2eq−1

20ohm−1cm2eq−1

30ohm−1cm2eq−1

25ohm−1cm2eq−1