Redox Reactions And Electrochemistry Question 731
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}} $
BETA
