1.

The conductivity of 0.20 M solution of KCI at 298 K is 0.0248 S cm^(-1).Calculate its molar conductivity.

Answer»

Solution :Applying the FOLLOWING relation and substituting the values, we get:
`Lambda_(m) = (k xx 1000)/("MOLARITY") = (0.0248 S CM^(-1) xx 1000 cm^(3) L^(-1))/(0.20 mol L^(-1)) = 124 S cm^(2) mol^(-1)`.


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