2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    \[\Lambda {{{}^\circ }_{m}}\] for \[NaCl,HCl\]and NaA are 126.4, 425.9 and \[100.5\,S\,c{{m}^{2}}\,mo{{l}^{-1}},\]respectively. If the conductivity of 0.001 M HA is \[5\times {{10}^{-5}}S\,c{{m}^{-1}},\]degree of dissociation of HA is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) 0.75                             

    B) 0.25

    C) 0.125               

    D)   0.50

    Correct Answer: C

    Solution :

    Given,\[\Lambda _{m}^{o}NaCl=126.4S\,c{{m}^{2}}mo{{l}^{-1}}\] \[\Lambda _{m}^{o}HCl=425.9S\,c{{m}^{2}}mo{{l}^{-1}}\] \[\Lambda _{m}^{{}}NaA=100.5S\,c{{m}^{2}}\,mo{{l}^{-1}}\] \[\Lambda _{m}^{o}HA=\Lambda _{m}^{o}HCl+\Lambda _{m}^{o}NaA-\Lambda _{m}^{o}NaCl\] \[=425.9+100.5-126.4\] \[=400S\,c{{m}^{2}}mo{{l}^{-1}}\] \[\Lambda _{m}^{{}}\frac{1000K}{C}=5\times {{10}^{-5}}\times \frac{1000}{0.001}=50\] \[\alpha =\frac{\Lambda _{m}^{{}}}{\Lambda _{m}^{o}}=\frac{50}{400}=0.125\]


You need to login to perform this action.
You will be redirected in 3 sec spinner