The resistance of conductivity cell containing 0.001MKCl solution at 298K is 1500 Ω . What is the cell constant if conductivity of 0.001MKCl solution at 298 K is 0.146× 10- 3 Scm- 1.

Question

The resistance of conductivity cell containing 0.001MKCl solution at 298K is 1500 Ω . What is the cell constant if conductivity of 0.001MKCl solution at 298 K is 0.146× 10- 3 Scm- 1. (A) 0.200 cm- 1 (B) 0.219 cm- 1 (C) 0.195 cm- 1 (D) 0.190 cm- 1

Tardigrade Admin · Accepted Answer

Cell constant =(c o n d u c t i v i t y/c o n d u c tan c e) = Conductivity × Resistance =0.146× 10- 3 Scm- 1× 1500 Ω =0.219 cm- 1

Q. The resistance of conductivity cell containing $0.001 M K Cl$ solution at $298 K$ is $1500 Ω$ . What is the cell constant if conductivity of $0.001 M K Cl$ solution at $298 K$ is $0.146 \times 1 0^{- 3} S c m^{- 1} .$

$0.200 c m^{- 1}$

$0.219 c m^{- 1}$

$0.195 c m^{- 1}$

$0.190 c m^{- 1}$

Cell constant $= \frac{co n d u c t i v i t y}{co n d u c t an ce}$
$=$ Conductivity $\times$ Resistance
$= 0.146 \times 1 0^{- 3} S c m^{- 1} \times 1500 Ω$
$= 0.219 c m^{- 1}$

Q. The resistance of conductivity cell containing 0.001MKCl solution at 298K is 1500Ω . What is the cell constant if conductivity of 0.001MKCl solution at 298K is 0.146×10−3Scm−1.

0.200cm−1

0.219cm−1

0.195cm−1

0.190cm−1