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The specific conductivity of a saturated solution of AgCl is `3.40xx10^(-6) ohm^(-1) cm^(-1)` at `25^(@)C`. If `lambda_(Ag^(+)=62.3 ohm^(-1) cm^(2) "mol"^(-1)` and `lambda_(Cl^(-))=67.7 ohm^(-1) cm^(2) "mol"^(-1)`, the solubility of AgC at `25^(@)C` is:A. `2.6xx10^(5)M`B. `3.73xx10^(-3)g//L`C. `3.7xx10^(-5)M`D. `2.6xx10^(-)g//L` |
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Answer» Correct Answer - A::B `A^(oo)=62.3+67.7=130Scm^(2)mol^(-1)` `rArrDelta^(oo)=(K1000)/(S)rArrS=(3.4xx10^(-6)xx1000)/(130)` `=2.6xx10^(-5)"molL"^(-1)` In `gL^(-1)` unit , solubility =`2.6xx10^(-5)xx143.5` `=3.7xx10^(-3)gL^(-1)` |
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