A) \[278\ S\ c{{m}^{2}}mo{{l}^{-1}}\]
B) \[176\ S\ c{{m}^{2}}mo{{l}^{-1}}\]
C) \[128\ S\ c{{m}^{2}}mo{{l}^{-1}}\]
D) \[302\ S\ c{{m}^{2}}mo{{l}^{-1}}\]
Correct Answer: C
Solution :
\[(126\ sc{{m}^{2}})\wedge _{NaCl}^{0}\ =\ \wedge _{N{{a}^{+}}}^{0}+\wedge _{C{{l}^{-}}}^{0}\] .....(1) \[(152\ sc{{m}^{2}})\wedge _{KBr}^{0}\ =\ \wedge _{{{K}^{+}}}^{0}+\wedge _{B{{r}^{-}}}^{0}\] .....(2) \[(150\ sc{{m}^{2}})\wedge _{KCl}^{0}\ =\ \wedge _{{{K}^{+}}}^{0}+\wedge _{C{{l}^{-}}}^{0}\] .....(3) By equation (1)+(2) ? (3) \[\because \]\[\wedge _{NaBr}^{0}\ =\ \wedge _{N{{a}^{+}}}^{0}+\wedge _{B{{r}^{-}}}^{0}\] \[=126+152-150=128\,\,Sc{{m}^{2}}\,mo{{l}^{-1}}\]
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