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The specific conductivity of a saturated solution of silver chloride is `2.30 xx 10^(-6) m ho cm^(-1) " at " 25^(@)C`. Calculate the solubility of silver chloride at `25^(@)C " if " lamda_(Ag^(+)) = 61.9 m ho cm^(2) mol^(-1) and lamda_(Cl^(-)) = 76.3 m ho cm^(2) mol^(-1)` |
Answer» Let the solubility of AgCl be S gram mole per litre Dilution `= (1000)/(S)` `Lamda_(AgCl)^(oo) = lamda_(Ag^(+)) + lamda_(Cl^(-))` `= 61.9 + 76.3` `= 138.2 m ho cm^(2) mol^(-1)` Sp. Conductivity `xx` Dilution `= Lamda_(AgCl)^(oo) = 138.2` `2.30 xx 10^(-6) xx (1000)/(S) = 138.2` `S = (2.30 xx 10^(-3))/(138.2) = 1.66 xx 10^(-5)` mol per litre `= 1.66 xx 10^(-5) xx 143.5 g L^(-1) = 2.382 xx 10^(-3) g L^(-1)` |
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