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The value of `Lambda_(eq)^(oo)` for `NH_(4)Cl, NaOH` and `NaCl` are `149.74, 248.1` and `126.4 Omega^(-1) cm^(2)" equiv"^(-1)`. The value of `Lambda_(eq)^(oo)` of `NH_(4)OH` isA. `371,44 Omega cm^(2)" equiv"^(-1)`B. `271,44 Omega cm^(2)" equiv"^(-1)`C. `71,44 Omega cm^(2)" equiv"^(-1)`D. data is insufficient to calculate it |
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Answer» Correct Answer - B `Lambda_(eq)^(oo) (NH_(4)OH)=Lambda_(eq)^(oo)(NH_(4)Cl)+ Lambda_(eq)^(oo) NaOH-Lambda_(eq)^(oo) (NaCl)` `=(149.74+248.1)-126.4` `=271.44 Omega cm^(2)" equiv"^(-1)` |
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