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Given the following molar conductivites at 25∘ C; HCl=426Ω−1 cm2 mol−1; NaCl=126Ω−1 cm2 mol−1, NaCro (sodium crotonate) =83Ω−1 cm2 mol−1. The conductivity of 0.001 mol/dm3 acid solution is 3.83×10−5Ω−1 cm−1. The dissociation constant of crotonic acid is

Answer»

Given the following molar conductivites at 25 C; HCl=426Ω1 cm2 mol1; NaCl=126Ω1 cm2 mol1, NaCro (sodium crotonate) =83Ω1 cm2 mol1. The conductivity of 0.001 mol/dm3 acid solution is 3.83×105Ω1 cm1. The dissociation constant of crotonic acid is




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