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The conductivity of a solution containing 1 gram of anhydrous BaCl_(2) in 200 cm^(3) of water has been found to be 0.0058 S cm^(-1). What are the molar conductivity and equivalent conductivity of the solution? (At. Wt. of Ba=137 and Cl=35.5) |
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Answer» Solution :We are given: Conductivity `(kappa)=0.0058" S "cm^(-1)` Molar MASS of `BaCl_(2)=137+2xx35.5=208" G "mol^(-1)` As 1 gram of `BaCl_(2)` is present in 200 `cm^(3)` of the solution, therefore, molar concentration (C) `=(1)/(208)xx(1)/(200)xx1000" mol "L^(-1)=0.0240" mol "L^(-1)` `therefore`Molar conductivity, `wedge_(m)=(kappaxx1000)/(c_(m))=(0.0058" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.0240" mol "L^(-1))=241.67" S "cm^(2)mol^(-1)` Further in case of `BaCl_(2)`, equivalent weight`=("Mol. wt")/(2)=(208)/(2)=104` `therefore` Concentration of the solution in gram equivalent per litre `(c_(EQ))=(1)/(104)xx(1)/(200)xx1000=0.0480` `therefore`Equivalent conductivity, `wedge_(eq)=(kappaxx1000)/(c_(eq))=(0.0058" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.0480" g "eqL^(-1))=120.83" S "cm^(2)eq^(-1)`. |
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