A) \[{{(\Lambda _{m}^{{}^\circ })}_{NaBr}}-(\Lambda _{m}^{{}^\circ }){{\,}_{NaCl}}={{(\Lambda _{m}^{{}^\circ })}_{kBr}}-{{(\Lambda _{m}^{{}^\circ })}_{KCl}}\]
B) \[{{(\Lambda _{m}^{{}^\circ })}_{{{H}_{2}}O}}={{(\Lambda _{m}^{{}^\circ })}_{\,HCl}}+{{(\Lambda _{m}^{{}^\circ })}_{NaO{{H}^{-}}}}{{(\Lambda _{m}^{{}^\circ })}_{NaCl}}\]
C) \[{{(\Lambda _{m}^{{}^\circ })}_{NaBr}}-(\Lambda _{m}^{{}^\circ }){{\,}_{NaI}}={{(\Lambda _{m}^{{}^\circ })}_{kBr}}-{{(\Lambda _{m}^{{}^\circ })}_{NaBr}}\]
D) \[{{(\Lambda _{m}^{{}^\circ })}_{KCl}}-(\Lambda _{m}^{{}^\circ }){{\,}_{NaCl}}={{(\Lambda _{m}^{{}^\circ })}_{kBr}}-{{(\Lambda _{m}^{{}^\circ })}_{NaBr}}\]
Correct Answer: C
Solution :
| [c] \[\Lambda _{m}^{{}^\circ }NaBr-\Lambda _{m}^{{}^\circ }NaI=\Lambda _{m}^{{}^\circ }(B{{r}^{-}})-\Lambda _{m}^{{}^\circ }({{I}^{-}})\] .......(1) |
| \[\Lambda _{m}^{{}^\circ }(KBr)-\Lambda _{m}^{{}^\circ }(NaBr)=\Lambda _{m}^{{}^\circ }({{K}^{+}})-\Lambda _{m}^{{}^\circ }(N{{a}^{+}})\] .....(2) |
| (1) and (2) are not equal. |
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