Redox Reactions And Electrochemistry Question 476
Question: The limiting equivalent conductivity of $ NaCl,~KCl $ and $ KBr $ are 126.5, 150.0 and $ 151.5Scm^{2}e{q^{-1}} $ , respectively. The limiting equivalent ionic conductivity for $ B{r^{-}} $ is $ 78,Scm^{2}e{q^{-1}} $ . The limiting equivalent ionic conductivity for $ N{a^{+}} $ ions would be:
Options:
A) 128
B) 125
C) 49
D) 50
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \Lambda _m^{\infty }(NaBr)=\Lambda _m^{\infty }(NaCl)+\Lambda _m^{\infty }KBr-\Lambda _m^{\infty }(KCl) $ $ \lambda _m^{\infty }(N{a^{+}})+\lambda _m^{\infty }(B{r^{-}})=126.5+151.5-150 $ $ \lambda _m^{\infty }(N{a^{+}})=50,Scm^{2}e{q^{-1}}. $
BETA
