1.

0.05 M NaOH solution offered a resistance of `3.1*6Omega` in a conductivity cell at 298 K. If the cell constant of the cell si `0*367cm^(-1),` calculate the molar conductivity of NaOH solution.

Answer» Specific conductance
(k)=C Conductance `xx` Cell constant …(1)
Conductance `=1/R=(1)/(31*6ohm)`
Cell constant `=0*367cm^(-)`
Substituting the values in equation (1), we get :
`k=(1)/(31*6ohm)xx0*367cm^(-)" "...(2)`
Also, molar conduuctivity,
`Delta_(m)=(kxx1000cm^(3)L^(-))/("Concentration"(mol//L))`
`=(1)/(31*6ohm)xx`
`(0*367cm^(-)1000cm^(3)//L)/(0*0.5mol//L)`
`=232*3ohm^(-)cm^(2)mol^(-1)`


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