Conductivity of, a solution is `6.23 xx 10^(-5) Omega^(-1) cm^(-1)` and its resistance is `13710 Omega`. If the electrode are 0.7 cm apart, calculate the cross-sectional area of electrode.

A conductivity cell dipped in 0.0005 M NaCl gives at 25^@C resistance of 13710 ohms.If the electrode of the conductivity cell are 0.7 cm apart and the area of cross section of the electrode is 0.82cm^2 ,what is the molar conductivity of the solution at 25^@C ?

The conductivity of 0.02M AgN03 at 25^@C is 2.428 xx 10^(-3)Omega^(-1) cm^(-1) . What is its molar conductivity?

Conductivity at 278 K of 0.15 mol. L^(-1) NaCl solution is 1.5 xx 10^(-2) ohm^(-1) cm^(-1) . Find molar conductance.

The conductivity of 0.1 mol L^(-1) KCl solution is 1.41 xx 10^(-3) S cm^(-1) . What is its molar conductivity (in S cm^(2) mol^(-1) ) ?

The specific conductance of 0.036 M KBr solution is 5.40 xx 10^(-4) S cm^(-1) . Find the molar conductivity of the solution.

A conductivity cell has two electrodes 50 mm apart and of cross section area 2.5 cm^2 . Find the cell constant.

A wire has length 10 m and resistance 20 Omega If the resistivity of the material of the wire is 6 times 10^(-7) Omega.m , find the area of cross section of the wire.

If the specific conductivity of 0.04 M electrolytic solution is 0.0649 Omega^(-1)m^(-1) , calculate its molar conductivity.

Specific conductance of 0.1M nitric acid is 6.3xx10^(-2)ohm^(-1)cm^(-1) . The molar conductance of the solution is:

Resistance of a conductivity cell filled with 0.1 M KC1 solution is 100 ohm. If the resistance of the same cell when filled with 0.02 M KC1 solution is520 ohms, calculate the conductivity and molar conductivity of 0.02 M KC1 solution. The conductivity of 0.1molL^(−1) KCl solution is 1.29×10 ^(−2) Ω ^(−1) cm ^(−1) .