A) 371.44
B) 271.44
C) 71.44
D) Cannot be predicted from given data
Correct Answer: B
Solution :
\[\Lambda _{eq}^{\infty }\,(N{{H}_{4}}OH)=\Lambda _{eq}^{\infty }\,(N{{H}_{4}}Cl)+\Lambda _{eq}^{\infty }\,\,(NaOH)\] \[-\Lambda _{eq}^{\infty }\,(NaCl)\] \[=(149.74+248.1-126.4)\] \[=271.44\,\,{{\Omega }^{-1}}c{{m}^{2}}e{{q}^{-1}}\]You need to login to perform this action.
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