1.

A conductivity cell has been calibrated with a 0.01M, 1:1 electrolytic solution (specific conductance (k = 1.25 xx 10^(-3)S cm^(-1)) in the cell and the measured resistance was 800 Omega at 25^@C. The cell constant is,

Answer»

`10^(-1) c m^(-1)`
`10^1 cm^(-1)`
`1 cm^(-1)`
`5.7 XX 10^(-12)`

SOLUTION :`R = rho.l/A`
Cell constant = `R/rho = k.R(1/(rho) = k) = 1.25 xx 10^(-3) mu^(-1)cm^(-1) xx 800 Omega = 1 cm^(-1)`.


Discussion

No Comment Found

Related InterviewSolutions