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A 0.05 M NH(4)OH solution offers the res...

A 0.05 M `NH_(4)OH` solution offers the resistance of 50 ohm to a conductivity cell at 298 K. If the cell constant is `0.50cm^(-1)` and molar conductance of `NH_(4)OH` at infinite dilution is `471.4ohm^(-1)cm^(2)mol^(-1)`, calculate :
Specific conductance

Text Solution

Verified by Experts

Specific conductance :
It is reciprocal of specific resistance.
`k=(1)/(rho)[because rho=(Ra)/(l)]`
`k=(1)/((Ra)/(l))=(1)/(R)xx(1)/(a)`
R = 50 ohm
`(l)/(a)` = cell constant = `0.50 cm^(-1)`
`k=(1)/(50)xx0.50`
`k=0.01 ohm^(-1)cm^(-1)`.
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