The equivalent conductance of a 0.2 N solution of an electrolyte was found to be 200 Ω -1 cm 2 eq -1. The cell constant of the cell is 2 cm -1. The resistance of the solution is

Question

The equivalent conductance of a 0.2 N solution of an electrolyte was found to be 200 Ω -1 cm 2 eq -1. The cell constant of the cell is 2 cm -1. The resistance of the solution is (A) 50 Ω  (B) 400 Ω  (C) 100 Ω  (D) None of these

Tardigrade Admin · Accepted Answer

Lambdav=200 Ω-1 cm 2 eq -1 C N =0.2 N , cm -1 K=( Lambdav × CN/1000)=(200 × 0.2/1000 × 10)=(1/25) Ω-1 cm -1 R=(1/K) ⋅ (l/a)=25 × 2=50 Ω .

Q. The equivalent conductance of a $0.2 N$ solution of an electrolyte was found to be $200 Ω^{- 1} c m^{2} e q^{- 1}$ . The cell constant of the cell is $2 c m^{- 1}$ . The resistance of the solution is

$50 Ω$

$400 Ω$

$100 Ω$

None of these

$Λ_{v} = 200 Ω^{- 1} c m^{2} e q^{- 1}$
$C_{N} = 0.2 N, c m^{- 1}$
$K = \frac{Λ _{v} \times C _{N}}{1000} = \frac{200 \times 0.2}{1000 \times 10} = \frac{1}{25} Ω^{- 1} c m^{- 1}$
$R = \frac{1}{K} \cdot \frac{l}{a} = 25 \times 2 = 50 Ω.$

Q. The equivalent conductance of a 0.2N solution of an electrolyte was found to be 200Ω−1cm2eq−1. The cell constant of the cell is 2cm−1. The resistance of the solution is

50Ω

400Ω

100Ω