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AMU Medical AMU Solved Paper-2015

  • question_answer
    0.01 m solution of \[KCl\] and \[BaC{{l}_{2}}\] are prepared in water. The freezing point of \[KCl\] found to be \[-{{4}^{o}}C\]. What will be the freezing point for \[BaC{{l}_{2}}\] solution assuming that both \[KCl\] and \[BaC{{l}_{2}}\] are completely ionised in solutions?

    A)  \[-{{3}^{o}}C\]                 

    B)  \[-{{4}^{o}}C\]

    C)  \[-{{5}^{o}}C\]                 

    D)  \[-{{6}^{o}}C\]

    Correct Answer: D

    Solution :

                    \[KCl{{K}^{+}}+C{{l}^{-}}\] (2 ions)                 \[BaC{{l}_{2}}B{{a}^{2+}}+2C{{l}^{-}}\] (3 ions) \[\because \] 2 ions shows depression in freezing point                 \[=-{{4}^{o}}C\]. \[\therefore \] 3 ions shows depression in freezing point                 \[=\frac{-4}{2}\times 3=-{{6}^{o}}C\]


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