1.

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is (100Omega). The conductivity of this solution is 1.29sm^(-1).Resistance of the same cell when filled with 0.2 M of the same solution is 520Omega . The molar conductivity of (0.02)M solution of the electrolyte will be :

Answer»

` 124xx10^(-4) sm^(2) MOL^(-1)`
`1240xx10^(-4) Sm^(2) mol^(-1) `
`1.242xx10^(-4)S m ^(2) mol^(-1)`
` 12.4xx10^(-4) sm ^(2) mol^(-1)`

SOLUTION : `K+ GxxG^ast`
For 0.1M solution `G^ast= (1.29)/(G) = 129m^(-1)`For 0.2M solution `K= Gxx(1)/(R ) = (129)/(520) sm^(-1)`
If we DILUTE 0.2 M solution by 10 times, its concentration becomes 0.02 M. Thus number of ions PER unit volume is `(1)/(10)`th of INITIAL one. So K for this solution is `(129)/(520xx10) SM^(-1)`


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