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KVPY Sample Paper KVPY Stream-SX Model Paper-18

  • question_answer
    Given the following molar conductivities at \[25{}^\circ C:,\] HCl, \[426\,\,{{\Omega }^{-\,1}}\,\,c{{m}^{2\text{ }}}mo{{l}^{-\,1}}\] NaCl, \[126\text{ }{{\Omega }^{-1}}c{{m}^{2\text{ }}}mo{{l}^{-1}};\] NaCl (sodium crotonate), \[\text{83 }{{\Omega }^{-1}}c{{m}^{2\text{ }}}mo{{l}^{-1}}.\] What is the ionization constant of crotonic acid? If the conductivity of a 0.001 M crotonic acid solution is \[3.83\text{ }\times \text{ 1}{{\text{0}}^{-5}}\text{ }{{\Omega }^{-1}}\text{ }c{{m}^{-1\text{ }}}?\]

    A) \[{{10}^{-5~~}}\]

    B) \[1.11\text{ }\times \text{ }{{10}^{-6}}\]          

    C) \[1.11\text{ }\times \text{ 1}{{0}^{-4~~}}\]

    D) 0.01

    Correct Answer: B

    Solution :

    \[{{\Lambda }_{m}}(HC)={{\Lambda }_{m}}(HCl)+{{\Lambda }_{m}}(NaC)-\Lambda m(NaCl)\]
    \[=426+83-126=383\,\,{{\Omega }^{-\,1}}\,c{{m}^{2}}\,mo{{l}^{-\,1}}\]
    Molar conductivity of HC
    \[{{\Lambda }_{m}}(HC)=\frac{k}{C}\]\[=\frac{3.83\times {{10}^{-5}}{{\Omega }^{-1}}c{{m}^{-1}}}{0.001}\times 1000\]\[=3.83\,{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}\]
    \[\alpha =\frac{{{\Lambda }_{m}}(HC)}{\Lambda _{m}^{\infty }(HC)}=\frac{38.3}{383}=0.1\]
    \[{{K}_{\alpha }}=\frac{C{{\alpha }^{2}}}{1-\alpha }=\frac{{{10}^{-\,3}}\times {{0.1}^{2}}}{(1-0.1)}=1.11\times {{10}^{-\,5}}\]


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