1.

The molar conductivity of 0.05 M BaCl_(2) solution at 25^(@)C is 223 Omega^(-1)"cm"^(2) mol^(-1). What is its conductivity?

Answer»


Solution :Molarconductivity `=^^_(m) =223 Omega^(-1) CM^(2) mol^(-1)`
Concetration= C= 0.05 M `BaCI_(2)`
CONDUCTIVITY = k = ?
`^^_(m) = (k xx 1000)/(C )`
`:. K =(^^_(m) xx C)/(1000)= (223 xx 0.05)/(1000)= 0.01115 Omega^(-1) cm^(-1)`


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