Redox Reactions And Electrochemistry Ques 330
Question: Molar conductance’s of $ BaC{l_{2,}} $ $ H_2SO_4 $ , and HCl at infinite dilutions are $ x_1 $ , $ x_2 $ , and $ x_3 $ , respectively. Equivalent conductance of $ BaSO_4 $ at infinite dilution will be:
Options:
A) $ \frac{[x_1+x_2-x_3]}{2} $
B) $ \frac{[x_1-x_2-x_3]}{2} $
C) $ 2(x_1+x_2-2x_3) $
D) $ \frac{[x_1+x_2-2x_3]}{2} $
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Answer:
Correct Answer: D
Solution:
[d] $ {\Lambda _{m,BaCl_2}}={\Lambda _{m}} _{Ba^{2+}}+2{\Lambda _{mCl^{-}}}x_1 $
$ {\Lambda_{m,H_2SO_4}}=2{\Lambda_{m{H^{+}}}}+{\Lambda_{m}}_{S{O^{-}}_4}x_2 $
$ {\Lambda_{mHCl}}={\Lambda_{m{H^{+}}}}+{\Lambda_{m}}_{C{l^{-}}}x_2]\times 2 $
$ \therefore {\wedge_{m,BaSO_4}}=(x_1+x_2-2x_3) $
$ \Rightarrow {\wedge_{eq.BaSO_4}}=\frac{{\wedge_{m,BaSO_4}}}{Totalcharge\text{(cation/anion)}} $
$ {\wedge_{eq.BaSO_4}}=\frac{[x_1+x_2-2x_3]}{2} $
