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The freezing point of an aqueous solution of KCN containing 0.189 mol $kg^{-1}$ was $-0.704^{\large\circ}$C. On adding 0.095 mol of $Hg(CN)_2$ , the freezing point of the solution became $-0.53^{\large\circ}$C. What will be new 'i' factor of the resulting? [Assuming that the $Hg(CN)_2$ and KCN produced the complex]

1 Answer

Hence (B) is the correct answer.
answered Jun 18, 2014 by sreemathi.v

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