1.

The specific conductivity of a solution containing 1.0 g of anhydrous BaCl_2 in 200 cm^3 of the solution has been found to be 0.0058 S cm^(-1). Calculate the molar conductivity of the solution. (Molecular wt. of BaCl_2 = 208).

Answer»


Solution :MOLARITY of `BaCl_(2)= (1 xx 1000)/(208 xx 200) = 0.024` M
Also, Normality of `BaCl_(2) = 0.024 xx 2 = 0.048` N
`(therefore N = M xx "VALENCY factor")`
Now, `wedge_(m) = k xx (1000)/C_(M) = (0.0058 xx 1000)/0.024`
`=241.67 S cm^(2) mol^(-1)`


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