1.

A 0-05 M sodium hydroxide offered a resistance of 31.6Omega in a conductivity cell at 298 K. Calculate the molar conductance (cell constant of cell =0367cm^(-1)).

Answer»

Solution :Specific conductance (K)
`=("CELL CONSTANT")/(R)`
`R=31.6Omega`
Cell constant `=0.367cm^(-1)` Specific conductance, (k)
`=(0.367)/(36.1)=0.0116Omegacm^(-1)`
Molar conductance, `Delta_(m)=(1000xxk)/(M)`
`=(1000xx0.0116)/(0.08)=232ohm^(-1)cm^(2)mol^(-1)`


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