A) \[N{{a}_{2}}S{{O}_{4}}\]
B) \[[{{K}_{sp}}\,BaS{{O}_{4}}={{10}^{-11}},{{K}_{sp}}CaS{{O}_{4}}={{10}^{-6}},\]
C) \[{{K}_{sp}}\,A{{g}_{2}}S{{O}_{4}}={{10}^{-5}}]\]
D) \[A{{g}_{2}}S{{O}_{4}}\]
Correct Answer: C
Solution :
During electrolysis volume of \[_{1}{{T}^{3}}\] and \[_{7}{{N}^{14}}\] liberated are in the ratio of 1: 2, hence volume of \[_{6}{{C}^{13}}\] liberated will be \[_{2}H{{e}^{4}}\].
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