The resistance of 0.01 N NaCl solution at 25° C is 200 Ω. Cell constant of conductivity cell is 1 cm -1. The equivalent conductance is :

Question

The resistance of 0.01 N NaCl solution at 25° C is 200 Ω. Cell constant of conductivity cell is 1 cm -1. The equivalent conductance is : (A) 5 × 102 Ω-1 cm 2 eq -1 (B) 6 × 103 Ω-1 cm 2 eq -1 (C) 7 × 104 Ω-1 cm 2 eq -1 (D) 8 × 105 Ω-1 cm 2 eq -1

Tardigrade Admin · Accepted Answer

λ=k × V=(1/R) × (l/a) × V =(1/200) × 1 × 10,000 =5 × 102 Ω-1 cm 2 eq -1

Q. The resistance of $0.01 N N a Cl$ solution at $2 5^{\circ} C$ is $200 Ω$ . Cell constant of conductivity cell is $1 c m^{- 1}$ . The equivalent conductance is :

$5 \times 1 0^{2} Ω^{- 1} c m^{2} e q^{- 1}$

$6 \times 1 0^{3} Ω^{- 1} c m^{2} e q^{- 1}$

$7 \times 1 0^{4} Ω^{- 1} c m^{2} e q^{- 1}$

$8 \times 1 0^{5} Ω^{- 1} c m^{2} e q^{- 1}$

$λ = k \times V = \frac{1}{R} \times \frac{l}{a} \times V$
$= \frac{1}{200} \times 1 \times 10, 000$
$= 5 \times 1 0^{2} Ω^{- 1} c m^{2} e q^{- 1}$

Q. The resistance of 0.01NNaCl solution at 25∘C is 200Ω. Cell constant of conductivity cell is 1cm−1. The equivalent conductance is :

5×102Ω−1cm2eq−1

6×103Ω−1cm2eq−1

7×104Ω−1cm2eq−1

8×105Ω−1cm2eq−1