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}} $

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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}} $



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