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JEE Main & Advanced AIEEE Solved Paper-2002

  • question_answer
    For an aqueous solution, freezing point is\[{{0.186}^{o}}C\]. Elevation of the boiling point of the same solution is (\[{{K}_{f}}={{1.86}^{o}}mo{{l}^{-1}}kg\] and \[{{K}_{b}}={{0512}^{o}}mo{{l}^{-1}}kg\])   AIEEE  Solved  Paper-2002

    A) \[{{0.186}^{o}}\]

    B)           \[{{0.0.512}^{o}}\]             

    C)            \[{{186}^{o}}\]    

    D)           \[{{5.12}^{o}}\]

    Correct Answer: B

    Solution :

    \[\Delta {{T}_{b}}=m{{K}_{b}}\]              \[\Delta {{T}_{f}}=m\,{{K}_{f}}\]              \[\frac{\Delta {{T}_{b}}}{\Delta {{T}_{f}}}=\frac{{{K}_{b}}}{{{K}_{f}}}=\frac{0512}{1.86}\]              \[\Delta {{T}_{b}}=\frac{0.512}{1.86}\times 0.186={{0.0512}^{o}}\]


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