A) \[{{\operatorname{S}}_{1}}<{{S}_{2}} <{{S}_{3}} <{{S}_{4}}\]
B) \[{{\operatorname{S}}_{1}} >{{S}_{3}} >{{S}_{2}} >{{S}_{4}}\]
C) \[{{\operatorname{S}}_{1}} >{{S}_{2}}={{S}_{3}} >{{S}_{4}}\]
D) \[{{\operatorname{S}}_{1}} >{{S}_{3}} >{{S}_{4}} >{{S}_{2}}\]
Correct Answer: B
Solution :
\[\operatorname{AgC}l\,\,\,A{{g}^{+}}\,\,+\,\,C{{l}^{-}}\] In \[{{\operatorname{CaCl}}_{2}},\,\,\,\,\underset{0.01}{\mathop{CaC{{l}_{2}}}}\,\,\,\,\underset{0.01}{\mathop{C{{a}^{2\,+}}}}\,+\,\,\underset{2\,\times \,0.01}{\mathop{2C{{l}^{-}}}}\,\] In \[\operatorname{NaC}l,\,\,\,\,\,\underset{0}{\mathop{NaCl}}\,\,\,=\,\,\underset{0.01}{\mathop{N{{a}^{+}}}}\,\,\,+\,\,\underset{0.01}{\mathop{C{{l}^{-}}}}\,\] In \[\operatorname{AgN}{{O}_{3}},\,\,\,\,\,\underset{0.05}{\mathop{AgN{{O}_{3}}}}\,\,\,=\,\,\underset{0.05}{\mathop{A{{g}^{+}}}}\,+\,\,\underset{0.05}{\mathop{C{{l}^{-}}}}\,\] Common ion effect is maximum in \[AgN{{O}_{3}}\] So, \[{{\operatorname{S}}_{1}} > {{S}_{3}} > {{S}_{2}} > {{S}_{4}}\]
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