1.

A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the cross sectional area of each electrode is 4.5 sq. cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as 15 Omega. Find the specific conductance of the solution.

Answer»

Solution :`k = 1/R(L/A)`
`k = 1/(15 omega) xx (1.5 xx 10^(-2) m)/(4.5 xx 10^(-4)m^2) = 2.22 SM^(-1)`
`l = 1.5 CM = 1.5 xx 10^2 m`
`A = 4.5 cm^2 xx (10^(-4)) m^2`
`R = 15 Omega` .


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