Redox Reactions And Electrochemistry Question 493
Given the ionic conductance of , $ {K^{+}} $ , and $ {Na^{+}} $ are 74, 50, and $ 73,cm^{2},ohm^{-1},eq^{-1} $ respectively. The equivalent conductance at infinite dilution of the salt is
Options:
A) $ 197,cm^{2},OH^{m^{-1}},e^{q^{-1}} $
B) $ 172,cm^{2},oh,m^{-1},e,q^{-1}$
C) $ 135.5,cm^{2}\cdot oh\cdot m^{-1}\cdot eq^{-1} $
D) $ 160.5,cm^{2}\cdot oh\cdot m^{-1}\cdot eq^{-1} $
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Answer:
Correct Answer: C
Solution:
Total charge = 2 × Number of equivalent of ion $ =\frac{Charge\ on\ the\ ion}{Total\ charge} $ $ \therefore Eq $ of $ =\frac{2}{2}=1 $ Eq of $ Na^{+}}=\frac{1}{2},Eqof,{K^{+}}=\frac{1}{2} $ $ =74+\frac{50}{2}+\frac{73}{2}=135.5oh{m^{-1}}cm^{2}eq^{-1}$
BETA
