At `25^(@)C` resistance of 0.01 (N) KCl and 0.01 (N) HCl solutions in the same conductivity cell are 150 ohm and 51.40 ohm, respectively. The specific conductance of the KCl solution at `25^(@)C` is `1.41xx10^(-3)" ohm"^(-1).cm"^(-1)`. Determine the molar conductivity of HCl solution `25^(@)C`.

Explanation

For KCl solution,
`"R = 150 ohm and "kappa =1.41xx10^(-3)"ohm"^(-1)."cm"^(-1)`.
Using the relation, `kappa=((1)/(R))xx(l)/(A)` for calculating the cell constant, we have
`1.41 xx10^(-3)=(1)/(50)xx(l)/(A)`
`"or, "(l)/(R)=(1.41xx10^(-3)xx150)"cm"^(-1)`
As the same cell is used in both cases, the value of cell constant will be the same.
In the case of HCl, `kappa _(HCl)=(1)/(R)xx(l)/(A)" "[because R=51.40 ohm]`
`=(1)/(51.40)xx(1.41xx10^(-3)xx150)="0.004114 ohm"^(-1)."cm"^(-1)`
Molar conductivity, `Lambda_(m)=kappa xx(1000)/(M)`. So the molar conductivity of the HCl soluiton,
`therefore" "Lambda_(m)=0.004114xx(1000)/(0.01)" [For HCl solution = 0.01 N = 0.01 M]"`
`="411.4 ohm"^(-1)."cm"^(2)."mol"^(-1)`