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The molar conductivity of 0.05 M BaCl(2)...

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?

Text Solution

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The correct Answer is:
Conductivity = k = 0.01115 `Omega^(-1) cm^(-1)`
Molar conductivity `=^^_(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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