WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer At a particular temperature, the ratio of equivalent   conductance   to   specific conductance of a 0.01 N NaCl solution is

    A)  \[10{{\,}^{5}}c{{m}^{3}}\]

    B)   \[10{{\,}^{3}}c{{m}^{3}}\]

    C)  \[10\,c{{m}^{3}}\]

    D)  \[10{{\,}^{5}}c{{m}^{2}}\]

    Correct Answer: A

    Solution :

     As we know that, Equivalent conductance, \[(\lambda )\] \[\text{=}\,\,\frac{\text{Specific}\,\text{consuctane}\,\text{(K) }\!\!\times\!\!\text{ 1000}}{\text{Concentration}}\] Or, \[\frac{\lambda }{K}=\frac{1000}{conc.}\] or, \[\frac{\lambda }{K}=\frac{1000\,{{\Omega }^{-1}}c{{m}^{2}}e{{q}^{-1}}}{0.01\,{{\Omega }^{-1}}\,c{{m}^{-1}}}\] (Given, conc. = 0.01 N) or, \[\frac{\lambda }{K}={{10}^{5}}\,c{{m}^{3}}\,e{{q}^{-1}}\]

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