JEE Main & Advanced Chemistry Equilibrium Question Bank Electrical conductors, Arrhenius theory and ostwald's dilution law

  • question_answer Vant hoff factor of \[BaC{{l}_{2}}\] of conc. \[0.01M\] is 1.98. Percentage dissociation of \[BaC{{l}_{2}}\] on this conc. Will be [Kerala CET 2005]

    A)                 49          

    B)                 69

    C)                 89          

    D)                 98

    E)                 100

    Correct Answer: A

    Solution :

                   \[BaC{{l}_{2}}\] ⇌ \[B{{a}^{2+}}\] + \[2C{{l}^{-}}\]                    Initially                   1             0             0                    After dissociation \[a-\alpha \]        \[\alpha \]         \[2\alpha \]                    Total = \[1-\alpha +\alpha +2\alpha =1+2\alpha \]                    \[\alpha =\frac{1.98-1}{\alpha }=\frac{0.98}{\alpha }=0.49\]                    for a mole  \[\alpha =0.49\]                                 For 0.01 mole   \[\alpha =\frac{0.49}{0.01}=49\]


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