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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Critical Thinking

  • question_answer
    A 0.004 M solution of \[N{{a}_{2}}S{{O}_{4}}\] is isotonic with a 0.010 M solution of glucose at same temperature. The apparent degree of disociation of \[N{{a}_{2}}S{{O}_{4}}\] is [IIT JEE Screening 2004]

    A)                 25%       

    B)                 50%

    C)                 75%       

    D)                 85%

    Correct Answer: C

    Solution :

               \[\underset{(0.004-x)}{\mathop{N{{a}_{2}}S{{O}_{4}}}}\,\] ⇌  \[\underset{2x}{\mathop{2N{{a}^{+}}}}\,+\underset{x}{\mathop{SO_{4}^{2-}}}\,\]                    Since both the solution are isotonic \[0.004+2x=0.01\]                    \[\therefore \ \ x=3\times {{10}^{-3}}\]                    \[\therefore \ \ \]Percent dissociation \[=\frac{3\times {{10}^{-3}}}{0.004}\times 100=75%\].


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