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The limiting molar conductance of sodium chloride, potassium chloride and potassium bromide are 126.45 , 149.86 and 151.92 ohm^(-1) cm^(2) mol^(-1) respectively . Calculate limiting molar ionic conductance of Na^(+) given that limiting molar ionic conductance of Br^(-) ion is 76.34 ohm^(-1) cm^(2) mol^(-1) |
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Answer» Solution :This can be calculated with the help of Kohlrausch's law because limiting MOLAR IONIC conductance are additive `Lambda_(m) ^(@) (KBr) = lambda_(m)^(@) (K^(+)) + lambda_(m)^(@) (Br^(-))` `Lambda_(m)^(@) (KBr) = 151.92 ohm^(-1) cm^(2) mol^(-1)`, `lambda_(m)^(@) (Br^(-)) = 76.34 ohm^(-1) cm^(2) mol^(-1)` `therefore lambda_(m)^(@) (K^(+)) = Lambda_(m)^(@) (KBr) - lambda_(m)^(@) (Br^(-))` = 151.92 - 76.34 =` 75.58 ohm^(-1) cm^(2) mol^(-1)` Now , `Lambda_(m)^(@) (KCl) = lambda_(m)^(@) (K^(+)) + lambda_(m)^(@) (Cl^(-))` `Lambda_(m)^(@) (KCl) = 149.86 ohm^(-1) cm^(2) mol^(-1)` , `lambda_(m)^(@) (K^(+)) = 75.58 ohm^(-1) cm^(2) mol^(-1)` `therefore lambda_(m)^(@) (Cl^(-)) = Lambda_(m)^(@) (KCl) - lambda_(m)^(@) (K^+)` `= 149.86 - 75.58` `= 74.28 ohm^(-1) cm^(2) mol^(-1)` Now `Lambda_(m)^(@) (NACL) = lambda_(m)^(@) (NA^(+)) + lambda_(m)^(@) (Cl^(-))` `Lambda_(m)^(@) (NaCl) = 126.45 ohm^(-1) cm^(2) mol^(-1)` Now `Lambda_(m)^(@) (NaCl) = lambda_(m)^(@) (Na^(+)) + lambda_(m)^(@) (Cl^(-))` `Lambda_(m)^(@) (NaCl) = 126.45 ohm^(-1) cm^(2) mol^(-1)` `lambda_(m)^(@) (Cl^(-)) = 74.28 ohm^(-1) cm^(2) mol^(-1)` `therefore lambda_(m)^(@) (Na^(+)) = Lambda_(m)^(@) (NaCl) - lambda_(m)^(@) (Cl^(-))` `=126.45 - 74.28` `=52.17 ohm^(-1) cm^(2) mol^(-1)` |
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