question_answer

Given the ionic conductance of , \[{{K}^{+}}\], and \[N{{a}^{+}}\] are 74, 50, and \[73\text{ }c{{m}^{2}}oh{{m}^{-1}}\] \[e{{q}^{-1}}\] respectively. The equivalent conductance at infinite dilution of the salt  is

A) \[197\text{ }c{{m}^{2}}oh{{m}^{-1}}e{{q}^{-1}}\]

B) \[172\,c{{m}^{2}}oh{{m}^{-1}}e{{q}^{-1}}\]

C) \[135.5\,c{{m}^{2}}oh{{m}^{-1}}e{{q}^{-1}}\]

D) \[160.5\,c{{m}^{2}}oh{{m}^{-1}}e{{q}^{-1}}\]

Correct Answer: C

Solution :

[c] Total charge =2 Number of equivalent of ion \[=\frac{Charge\text{ }on\text{ }the\text{ }ion}{Total\text{ }charge}\] \[\therefore Eq\] of  \[=\frac{2}{2}=1\] Eq of \[N{{a}^{+}}=\frac{1}{2},\text{ }Eq\text{ }of\,{{K}^{+}}=\frac{1}{2}\] \[=74+\frac{50}{2}+\frac{73}{2}=135.5\text{ }oh{{m}^{-1}}c{{m}^{2}}e{{q}^{-1}}\]