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JEE Main & Advanced Sample Paper JEE Main Sample Paper-13

  • question_answer
    The concentration of \[O{{H}^{-}}\] ion in a solution left after mixing 100 mL of \[0.1\,\,M\,MgC{{l}_{2}}\] and 100 mL of \[0.2\,M\,NaOH\] \[({{K}_{sp}}\,[Mg{{(OH)}_{2}}]=1.2\times {{10}^{-11}})\] is

    A)  \[2.8\times {{10}^{-3}}\]                             

    B)  \[2.8\times {{10}^{-4}}\]

    C)  \[2\times {{10}^{-7}}\]                 

    D)  \[{{10}^{-7}}\]

    Correct Answer: B

    Solution :

     \[b(ac+b)+a(bc+a)-(1-{{c}^{2}})=0\] M moles 10         20           0                              0 0              0              20                                           0 i.e., almost all \[b(ac+b)+a(bc+a)-(1-{{c}^{2}})=0\] gets precipitated, Hence,  the remaining \[\Rightarrow \] and \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc=1\] concentration can be determined from \[(2\pi ct/\lambda )\] value of \[(2\pi x/\lambda )\]. \[x\] \[\lambda \] \[\left[ \frac{2\pi Ct}{\lambda } \right]=\,\left[ \frac{2\pi x}{\lambda } \right]=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] \[\frac{2\pi C}{\lambda }=\frac{2\pi x}{\lambda t}\]    \[\frac{x}{\lambda }\]


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