The specific conductance of 0.1 M NaCl solution is 3.2 × 10-3 S cm -1 . The resistance of the solution contained in the cell is found to be 125 Ω . The cell constant (in cm -1 ) will be equal to

Question

The specific conductance of 0.1 M NaCl solution is 3.2 × 10-3 S cm -1 . The resistance of the solution contained in the cell is found to be 125 Ω . The cell constant (in cm -1 ) will be equal to (A) 0.4 (B) 2.5 (C) 3.9 (D) 0.9

Tardigrade Admin · Accepted Answer

kappa=(1/R) × (1/a) ⇒ Cell constant, (l/a)= kappa × R =3.2 × 10-3 × 125=0.4 cm -1

Q. The specific conductance of $0.1 M N a Cl$ solution is $3.2 \times 1 0^{- 3} S c m^{- 1} .$ The resistance of the solution contained in the cell is found to be $125 Ω.$ The cell constant (in $c m^{- 1}$ ) will be equal to

0.4

2.5

3.9

0.9

$κ = \frac{1}{R} \times \frac{1}{a}$
$\Rightarrow$ Cell constant, $\frac{l}{a} = κ \times R$
$= 3.2 \times 1 0^{- 3} \times 125 = 0.4 c m^{- 1}$

Q. The specific conductance of 0.1MNaCl solution is 3.2×10−3Scm−1. The resistance of the solution contained in the cell is found to be 125Ω. The cell constant (in cm−1 ) will be equal to

0.4

2.5

3.9

0.9

κ=R1​×a1​ ⇒ Cell constant, al​=κ×R =3.2×10−3×125=0.4cm−1

Q. The specific conductance of $0.1 M N a Cl$ solution is $3.2 \times 1 0^{- 3} S c m^{- 1} .$ The resistance of the solution contained in the cell is found to be $125 Ω.$ The cell constant (in $c m^{- 1}$ ) will be equal to

$κ = \frac{1}{R} \times \frac{1}{a}$
$\Rightarrow$ Cell constant, $\frac{l}{a} = κ \times R$
$= 3.2 \times 1 0^{- 3} \times 125 = 0.4 c m^{- 1}$