A) 203
B) 279
C) 101.5
D) 139.5
Correct Answer: B
Solution :
[b] Equivalent conductivity of \[BaC{{l}_{2}}\] \[\Lambda _{m}^{\infty }(BaC{{l}_{2}})=\lambda _{m}^{\infty }(B{{a}^{2+}})+2\lambda _{m}^{\infty }(C{{l}^{-}})\] \[=127+2\times 76=279\text{ }oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}.\]You need to login to perform this action.
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