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At a definite temperature, with the same...

At a definite temperature, with the same conductivity cell, the specific conductance of `"0.01 N KCl"` and `"0.01 N NaCl"` solutions are `"y ohm"^(-1)."cm"^(-1) and "z ohm"^(-1)."cm"^(-1)` respectively. If the conductance of KCl solution is `"x ohm"^(-1)`, what will be the conductance of `NaCl` solution?

Text Solution

Verified by Experts

We know, specific conductance, `kappa =Lxx(l)/(A),`
As given, `kappa_(KCl)="y ohm"^(-1)."cm"^(-1)`
`kappa_(NaCl)="z ohm"^(-1)."cm"^(-1)`
`L_(KCl)="x ohm"^(-1)`
`therefore" For a particular conductivity cell, "(l)/(A)` is fixed.
`(kappa_(KCl))/(kappa_(NaCl))=(L_(KCl))/(L_(NaCl))`
`or, (y)/(z)=(x)/(L_(NaCl)) or, L_(NaCl)=(zx)/(y)`
`therefore" conductance of NaCl solution will be "(zx)/(y)" ohm"^(-1)`.
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