A) \[1505\,kJ\,mo{{l}^{-1}}\]
B) \[1310\,kJ\,mo{{l}^{-1}}\]
C) \[1608\,kJ\,mo{{l}^{-1}}\]
D) None
Correct Answer: B
Solution :
[b] \[\text{I}\text{.E}\text{.=}\frac{\text{2}\text{.18 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-18}}}\text{J}}{\text{1atom}}\text{ }\!\!\times\!\!\text{ }\frac{\text{6}\text{.02 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}\text{atom}}{\text{mole}}\]\[\text{=1}\text{.31 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\text{J mol}{{\text{e}}^{-1}}\text{=1310 KJ mol}{{\text{e}}^{\text{-1}}}\]
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