1.

What will be the value of a for a 0.001 M aqueous NH_(3) solution ? [K_(b)=1.6xx 10^(-5) and Lambda_(0)=2.38xx10^(-2) ohm^(-1) m^(2) mol^(-1)]

Answer»


Solution :`K_(b)=Calpha^(2)`
`1.6xx10^(-5)=0.001xxalpha^(2)`
`alpha=0.126`
`alpha=Lambda_(m)^(C)/Lambda_(m)^(oo),`
(`Lambda_(m)^(c)=` MOLAR conductance at concentration 'c' and `Lambda_(m)^(oo)=` molar conductance at infinite DILUTION)
`0.126=Lambda_(m)^(c)/(2.38xx10^(-2))`
`Lambda_(m)^(c)=2.998xx10^(-3) ohm^(-1) cm^(2) mol^(-1)`


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