Redox Reactions And Electrochemistry Ques 626
Question: $ {\Lambda_{ClCH_2COONa}}=224,oh{m^{-1}}cm^{2}g,e{q^{-1}}, $
$ {\Lambda_{NaCl}}=38.2,oh{m^{-1}}cm^{2}g,e{q^{-1}}, $ $ {\Lambda_{HCl}}=203,oh{m^{-1}}cm^{2}ge{q^{-1}}, $ What is the value of $ {\Lambda_{CICH_2COOH}} $
Options:
A) $ 288.5,oh{m^{-1}}cm^{2},g,e{q^{-1}} $
B) $ 289.5,oh{m^{-1}}cm^{2}ge{q^{-1}} $
C) $ 388.8,oh{m^{-1}}cm^{2}g,e{q^{-1}} $
D) $ 59.5,oh{m^{-1}}cm^{2}g,e{q^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ ClCH_2COONa+HCl\to ClCH_2OOH+NaCl $ $ {\Lambda_{ClCH_2COONa}}+{\Lambda_{HCl}}={\Lambda_{ClCH_2COOH}}+{\Lambda_{NaCl}} $ $ 224+203={\Lambda_{ClCH_2COOH}}+38.2 $ $ {\Lambda_{CICH_2COOH}}=427-38.2 $ $ =388.8,oh{m^{-1}}cm^{2}g,e{q^{-1}} $
