Two platinum electrodes, each of area 10 sq. cm are fixed 1.5 cm apart in a conductivity cell. The cell contain 0.05 N solution of a salt with resistance 50 ohm. What is the equivalent conductivity of the salt solution?

Question

Two platinum electrodes, each of area 10 sq. cm are fixed 1.5 cm apart in a conductivity cell. The cell contain 0.05 N solution of a salt with resistance 50 ohm. What is the equivalent conductivity of the salt solution? (A) 66.6 S cm 2 eq -1 (B) 15 S cm 2 eq -1 (C) 60 S cm 2 eq -1 (D) 6.66 S cm 2 eq -1

Tardigrade Admin · Accepted Answer

K =(1/R) × (l/a)=(1/50) × (1.5/10) Eq. conductance = kappa × (1000/ text Normality ) =(1/50) × (1.5/10) × (1000/0.05)=60 S cm 2 eq -1

Q. Two platinum electrodes, each of area $10 s q . c m$ are fixed $1.5 c m$ apart in a conductivity cell. The cell contain $0.05 N$ solution of a salt with resistance $50 o hm$ . What is the equivalent conductivity of the salt solution?

$66.6 S c m^{2} e q^{- 1}$

$15 S c m^{2} e q^{- 1}$

$60 S c m^{2} e q^{- 1}$

$6.66 S c m^{2} e q^{- 1}$

$K = \frac{1}{R} \times \frac{l}{a} = \frac{1}{50} \times \frac{1.5}{10}$

Eq. conductance $= κ \times \frac{1000}{Normality}$

$= \frac{1}{50} \times \frac{1.5}{10} \times \frac{1000}{0.05} = 60 S c m^{2} e q^{- 1}$

Q. Two platinum electrodes, each of area10sq.cm are fixed 1.5cm apart in a conductivity cell. The cell contain 0.05N solution of a salt with resistance 50ohm. What is the equivalent conductivity of the salt solution?

66.6Scm2eq−1

15Scm2eq−1

60Scm2eq−1

6.66Scm2eq−1