JCECE Engineering JCECE Engineering Solved Paper-2012

  • question_answer The resistance of \[1\,\,N\] solution of acetic acid is\[250\,\Omega \], when measured in a cell having a cell constant of\[1.15\,\,c{{m}^{-1}}\]. The equivalent conductance (in\[{{\Omega }^{-1}}c{{m}^{2}}equi{{v}^{-1}})\] of \[1\,\,N\] acetic acid is

    A) \[2.3\]                                  

    B) \[4.6\]

    C) \[9.2\]                                  

    D)  \[18.4\]

    Correct Answer: B

    Solution :

    Equivalent conductivity\[({{\Lambda }_{eq}})=\frac{\kappa \times 1000}{c}\]                 \[\kappa =\frac{\text{cell}\,\,\text{constant}}{\text{resistance}}=\frac{1.15}{250}S\,\,c{{m}^{-1}}\]                 \[{{\Lambda }_{eq}}=\frac{1.15\times 1000}{250\times 1}=4.6{{\Omega }^{-1}}c{{m}^{2}}equi{{v}^{-1}}\]

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