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) \[\frac{[{{x}_{1}}+{{x}_{2}}-2{{x}_{3}}]}{2}\]
Correct Answer: D
Solution :
[d] \[{{\Lambda }_{m,BaC{{l}_{2}}}}={{\Lambda }_{m}}_{B{{a}^{2+}}}+2{{\Lambda }_{mC{{l}^{-}}}}{{x}_{1}}\] \[{{\Lambda }_{m,{{H}_{2}}S{{O}_{4}}}}=2{{\Lambda }_{m{{H}^{+}}}}+{{\Lambda }_{m}}_{S{{O}^{-}}_{4}}{{x}_{2}}\] \[{{\Lambda }_{mHCl}}={{\Lambda }_{m{{H}^{+}}}}+{{\Lambda }_{m}}_{C{{l}^{-}}}{{x}_{2}}]\times 2\] \[\therefore {{\wedge }_{m,BaS{{O}_{4}}}}=({{x}_{1}}+{{x}_{2}}-2{{x}_{3}})\] \[\Rightarrow {{\wedge }_{eq.BaS{{O}_{4}}}}=\frac{{{\wedge }_{m,BaS{{O}_{4}}}}}{\text{Total}\text{charge}\text{(cation/anion)}}\] \[{{\wedge }_{eq.BaS{{O}_{4}}}}=\frac{[{{x}_{1}}+{{x}_{2}}-2{{x}_{3}}]}{2}\]
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