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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    0.01 M solution of\[KCl\]and\[BaC{{l}_{2}}\]are prepared in water. The freezing points of \[KCl\]is found to be\[-2{}^\circ C\]. What is the freezing point of \[BaC{{l}_{2}}\] solution assuming both\[KCl\] and\[BaC{{l}_{2}}\]to be completely ionized?

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

    B)         \[+3{}^\circ C\]

    C)  \[-2{}^\circ C\]                

    D)         \[-4{}^\circ C\]

    E)  \[5{}^\circ C\]

    Correct Answer: A

    Solution :

    \[i\]for\[KCl=2,i\]for\[BaC{{l}_{2}}=3\] \[\because \]     \[\Delta {{T}_{f}}\propto i\]                 \[\frac{\Delta {{T}_{f}}(KCl)}{\Delta {{T}_{f}}(BaC{{l}_{2}})}=\frac{2}{3}\] \[\Delta {{T}_{f}}(BaC{{l}_{2}})=\frac{3}{2}\times 2=3{}^\circ C\] \[\therefore \]freezing point of \[KCl=-3{}^\circ C\]


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