Redox Reactions And Electrochemistry Ques 84

Question: The limiting molar conductivities $ {{\wedge }^{0}} $ for NaCl, KBr and KCl are 126, 152 and 150 $ S\ cm^{2}mo{l^{-1}} $ respectively. The $ {{\wedge }^{0}} $ for NaBr is [AIEEE 2004]

Options:

A) $ 278\ S\ cm^{2}mo{l^{-1}} $

B) $ 176\ S\ cm^{2}mo{l^{-1}} $

C) $ 128\ S\ cm^{2}mo{l^{-1}} $

D) $ 302\ S\ cm^{2}mo{l^{-1}} $

Show Answer

Answer:

Correct Answer: C

Solution:

$ (126\ scm^{2})\wedge _NaCl^{0}\ =\ \wedge _{N{a^{+}}}^{0}+\wedge _{C{l^{-}}}^{0} $ …..(1)

$ (152\ scm^{2})\wedge _KBr^{0}\ =\ \wedge _{{K^{+}}}^{0}+\wedge _{B{r^{-}}}^{0} $ …..(2)

$ (150\ scm^{2})\wedge _KCl^{0}\ =\ \wedge _{{K^{+}}}^{0}+\wedge _{C{l^{-}}}^{0} $ …..(3)

By equation (1)+(2) - (3)

$ \because $ $ \wedge _{NaBr}^{0}\ =\ \wedge _{N{a^{+}}}^{0}+\wedge _{B{r^{-}}}^{0} $

$ =126+152-150=128Scm^{2},mo{l^{-1}} $



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