A) 0.10
B) 0.02
C) 0.15
D) 0.03
E) 0.20
Correct Answer: B
Solution :
According to Kohlrauschs law, \[\wedge {}^\circ \]for\[C{{H}_{3}}COOH=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{{{H}^{+}}}^{o}\] \[\wedge {}^\circ \]for \[NaCl=\lambda _{N{{a}^{+}}}^{o}+\lambda _{C{{l}^{-}}}^{o}\] \[=125\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}}\] ... (i) \[\wedge {}^\circ \]for \[HCl=\lambda _{{{H}^{+}}}^{o}+\lambda _{C{{l}^{-}}}^{o}\] \[=425\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}}\] ...(ii) \[\wedge {}^\circ \]for \[C{{H}_{3}}COONa=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{N{{a}^{+}}}^{o}\] \[=90\,mho\,c{{m}^{2}}mo{{l}^{-1}}\] ...(iii) Adding Eqs. (ii) and (iii) and subtracting (i), we get \[\lambda _{{{H}^{+}}}^{o}+\lambda _{C{{l}^{-}}}^{{}^\circ }+\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{N{{a}^{+}}}^{o}-\lambda _{Na}^{o}-\lambda _{C{{l}^{-}}}^{o}\] \[=425+90-125\] \[=390\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}}\] Or \[\wedge _{C{{H}_{3}}COOH}^{o}=\lambda _{C{{H}_{3}}CO{{O}^{-}}}^{o}+\lambda _{{{H}^{+}}}^{o}\] \[=390\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}}\] Thus, the molar conductivity of\[C{{H}_{3}}COOH\]at infinite dilution \[\wedge {}^\circ =390\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}}\] The molar conductivity of\[0.1\text{ }M\text{ }C{{H}_{3}}COOH\] solution\[(\wedge _{m}^{c})\] \[=7.8\text{ }mho\text{ }c{{m}^{2}}\text{ }mo{{l}^{-1}}\] Degree of dissociation \[(\alpha )=\frac{\wedge _{m}^{c}}{\wedge _{m}^{o}}=\frac{7.8}{390}=0.02\]
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