In a conductivity cell, the two platinum electrodes, each of area 10 cm 2, are fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, find the equivalent conductance of the salt solution is

Question

In a conductivity cell, the two platinum electrodes, each of area 10 cm 2, are fixed 1.5 cm apart. The cell contained 0.05 N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, find the equivalent conductance of the salt solution is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Lambda eq =(1000/ c ) × (l/ a ) × G =(1000 × 10-6 m 3 L -1/0.05 eq L -1) × (1.5 × 10-2 m /5.0 × 10-4 m 2) × (1/50 ohm ) =0.012 S m 2 eq -1

Q. In a conductivity cell, the two platinum electrodes, each of area 10cm2, are fixed 1.5cm apart. The cell contained 0.05N solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, find the equivalent conductance of the salt solution is

Λeq​=c1000​×al​×G =0.05eqL−11000×10−6m3L−1​×5.0×10−4m21.5×10−2m​×50ohm1​ =0.012Sm2eq−1

$Λ_{e q} = \frac{1000}{c} \times \frac{l}{a} \times G$

$= \frac{1000 \times 1 0 ^{- 6} m ^{3} L ^{- 1}}{0.05 e q L ^{- 1}} \times \frac{1.5 \times 1 0 ^{- 2} m}{5.0 \times 1 0 ^{- 4} m ^{2}} \times \frac{1}{50 o hm}$

$= 0.012 S m^{2} e q^{- 1}$